Count Even Numbers in Array Question: Count how many even numbers are present in an array. Code: #include <iostream> using namespace std; int main() { int arr[] = {1, 4, 7, 8, 10}; int n = 5, count = 0; for (int i = 0; i < n; i++) { // TODO: Fill here } cout << count; }

if (arr[i] % 2 == 0) {count++};

Find Maximum Element in Array Question: Find the largest element in an array. Code: #include <iostream> using namespace std; int main() { int arr[] = {3, 8, 2, 10, 7}; int n = 5; int maxVal = arr[0]; for (int i = 1; i < n; i++) { // TODO: Fill here } cout << maxVal; }

if (arr[i] > maxVal) maxVal = arr[i];

if (arr[i] > maxVal) { maxVal = arr[i]};

Check Prime Question: Check whether a number n is prime. Code: #include <iostream> #include <cmath> using namespace std; int main() { int n = 17; bool isPrime = true; for (int i = 2; i <= sqrt(n); i++) { // TODO: Fill here } cout << (isPrime ? "Prime" : "Not Prime"); }

if (n % i == 0) { isPrime = false; break; }

Count Vowels in String Question: Count the number of vowels in a given string. Code: #include <iostream> #include <string> using namespace std; int main() { string str = "placement"; int count = 0; for (char c : str) { // TODO: Fill here } cout << count; }

if (c=='a'||c=='e'||c=='i'||c=='o'||c=='u'||c=='A'||c=='E'||c=='I'||c=='O'||c=='U') count++;

Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation . Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval(). Example 1: Input: s = "1 + 1" Output: 2 Example 2: Input: s = " 2-1 + 2 " Output: 3 Example 3: Input: s = "(1+(4+5+2)-3)+(6+8)" Output: 23 Constraints: 1 <= s.length <= 3 * 10 5 s consists of digits, '+', '-', '(', ')', and ' '. s represents a valid expression. '+' is not used as a unary operation (i.e., "+1" and "+(2 + 3)" is invalid). '-' could be used as a unary operation (i.e., "-1" and "-(2 + 3)" is valid). There will be no two consecutive operators in the input. Every number and running calculation will fit in a signed 32-bit integer.

for (int i = 0; i < s.size(); i++) { if (isdigit(s[i])) { int num = 0; while (i < s.size() && isdigit(s[i])) { num = num * 10 + (s[i] - '0'); i++; } i--; result += sign * num; } else if (s[i] == '+') sign = 1; else if (s[i] == '-') sign = -1; else if (s[i] == '(') { st.push(result); st.push(sign); result = 0; sign = 1; } else if (s[i] == ')') { result = st.top () * result; st.pop(); result += st.top (); st.pop(); } }

for (int i = 0; i < s.size(); i++) { if (isdigit(s[i])) { int num = 0; while (i < s.size() && isdigit(s[i])) { num = num * 10 + (s[i] - '0'); i++; } i--; result += sign * num; } else if (s[i] == '+') sign = 1; else if (s[i] == '-') sign = -1; else if (s[i] == '(') { st.push(result); st.push(sign); result = 0; sign = 1; } else if (s[i] == ')') { result = result * sign; // wrong result += st.top (); st.pop(); st.pop(); // discards sign incorrectly } }

for (int i = 0; i < s.size(); i++) { if (isdigit(s[i])) { int num = 0; while (i < s.size() && isdigit(s[i])) { num = num * 10 + (s[i] - '0'); i++; } i--; result += sign * num; } else if (s[i] == '+') sign = 1; else if (s[i] == '-') sign = -1; else if (s[i] == '(') { st.push(result); st.push(sign); result = 0; // missing sign reset here } else if (s[i] == ')') { result = st.top () * result; st.pop(); result += st.top (); st.pop(); } }

for (int i = 0; i < s.size(); i++) { if (isdigit(s[i])) { int num = 0; while (i < s.size() && isdigit(s[i])) { num = num * 10 + (s[i] - '0'); i++; } i--; result += sign * num; } else if (s[i] == '+') sign = 1; else if (s[i] == '-') sign = -1; else if (s[i] == '(') { st.push(sign); // wrong: sign pushed first st.push(result); result = 0; sign = 1; } else if (s[i] == ')') { result = st.top () * result; st.pop(); result += st.top (); st.pop(); } }

You are given an array of unique strings words where words[i] is six letters long. One word of words was chosen as a secret word. You are also given the helper object Master. You may call Master.guess(word) where word is a six-letter-long string, and it must be from words. Master.guess(word) returns: -1 if word is not from words, or an integer representing the number of exact matches (value and position) of your guess to the secret word. There is a parameter allowedGuesses for each test case where allowedGuesses is the maximum number of times you can call Master.guess(word). For each test case, you should call Master.guess with the secret word without exceeding the maximum number of allowed guesses. You will get: "Either you took too many guesses, or you did not find the secret word." if you called Master.guess more than allowedGuesses times or if you did not call Master.guess with the secret word, or "You guessed the secret word correctly." if you called Master.guess with the secret word with the number of calls to Master.guess less than or equal to allowedGuesses. The test cases are generated such that you can guess the secret word with a reasonable strategy (other than using the bruteforce method). Example 1: Input: secret = "acckzz", words = ["acckzz","ccbazz","eiowzz","abcczz"], allowedGuesses = 10 Output: You guessed the secret word correctly. Explanation: master.guess("aaaaaa") returns -1, because "aaaaaa" is not in wordlist. master.guess("acckzz") returns 6, because "acckzz" is secret and has all 6 matches. master.guess("ccbazz") returns 3, because "ccbazz" has 3 matches. master.guess("eiowzz") returns 2, because "eiowzz" has 2 matches. master.guess("abcczz") returns 4, because "abcczz" has 4 matches. We made 5 calls to master.guess, and one of them was the secret, so we pass the test case. Example 2: Input: secret = "hamada", words = ["hamada","khaled"], allowedGuesses = 10 Output: You guessed the secret word correctly. Explanation: Since there are two words, you can guess both. Constraints: 1 <= words.length <= 100 words[i].length == 6 words[i] consist of lowercase English letters. All the strings of wordlist are unique . secret exists in words. 10 <= allowedGuesses <= 30

class Solution { public: void findSecretWord(vector<string>& words, Master& master) { for (int t = 0; t < 10; ++t) { string guessWord = words[rand() % words.size()]; int matches = master.guess(guessWord); vector<string> filtered; for (auto &w : words) { if (matchCount(guessWord, w) == matches) { filtered.push_back(w); } } words = filtered; } } int matchCount(string &a, string &b) { int cnt = 0; for (int i = 0; i < 6; i++) if (a[i] == b[i]) cnt++; return cnt; } };

class Solution { public: void findSecretWord(vector<string>& words, Master& master) { for (int t = 0; t < 10; ++t) { string guessWord = words[rand() % words.size()]; int matches = master.guess(guessWord); vector<string> filtered; for (auto &w : words) { if (matchCount(guessWord, w) != matches) { filtered.push_back(w); } } words = filtered; } } int matchCount(string &a, string &b) { int cnt = 0; for (int i = 0; i < 6; i++) if (a[i] == b[i]) cnt++; return cnt; } };

class Solution { public: void findSecretWord(vector<string>& words, Master& master) { for (int t = 0; t < 10; ++t) { string guessWord = words[rand() % words.size()]; int matches = master.guess(guessWord); vector<string> filtered; for (auto &w : words) { if (matchCount(guessWord, w) == matches) { filtered.push_back(w); } } words = filtered; } } int matchCount(string &a, string &b) { int cnt = 0; for (int i = 0; i < 6; i++) if (a[i] == b[5 - i]) cnt++; return cnt; } };

class Solution { public: void findSecretWord(vector<string>& words, Master& master) { for (int t = 0; t < 10; ++t) { string guessWord = words[rand() % words.size()]; master.guess(guessWord); words.pop_back(); } } };

There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous. You can feed the pigs according to these steps: Choose some live pigs to feed. For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time. Each pig can feed from any number of buckets, and each bucket can be fed from by any number of pigs. Wait for minutesToDie minutes. You may not feed any other pigs during this time. After minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive. Repeat this process until you run out of time. Given buckets, minutesToDie, and minutesToTest, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time . Example 1: Input: buckets = 4, minutesToDie = 15, minutesToTest = 15 Output: 2 Explanation: We can determine the poisonous bucket as follows: At time 0, feed the first pig buckets 1 and 2, and feed the second pig buckets 2 and 3. At time 15, there are 4 possible outcomes: - If only the first pig dies, then bucket 1 must be poisonous. - If only the second pig dies, then bucket 3 must be poisonous. - If both pigs die, then bucket 2 must be poisonous. - If neither pig dies, then bucket 4 must be poisonous. Example 2: Input: buckets = 4, minutesToDie = 15, minutesToTest = 30 Output: 2 Explanation: We can determine the poisonous bucket as follows: At time 0, feed the first pig bucket 1, and feed the second pig bucket 2. At time 15, there are 2 possible outcomes: - If either pig dies, then the poisonous bucket is the one it was fed. - If neither pig dies, then feed the first pig bucket 3, and feed the second pig bucket 4. At time 30, one of the two pigs must die, and the poisonous bucket is the one it was fed. Constraints: 1 <= buckets <= 1000 1 <= minutesToDie <= minutesToTest <= 100

#include <bits/stdc++.h> using namespace std; int poorPigs(int buckets, int minutesToDie, int minutesToTest) { int trials = minutesToTest / minutesToDie + 1; int pigs = 0; while (pow(trials, pigs) < buckets) { pigs++; } return pigs; } int main() { int buckets = 4, minutesToDie = 15, minutesToTest = 15; cout << poorPigs(buckets, minutesToDie, minutesToTest) << endl; }

#include <bits/stdc++.h> using namespace std; int poorPigs(int buckets, int minutesToDie, int minutesToTest) { int trials = minutesToTest / minutesToDie; int pigs = 0; while (pow(trials, pigs) < buckets) { pigs++; } return pigs; } int main() { int buckets = 4, minutesToDie = 15, minutesToTest = 15; cout << poorPigs(buckets, minutesToDie, minutesToTest) << endl; }

#include <bits/stdc++.h> using namespace std; int poorPigs(int buckets, int minutesToDie, int minutesToTest) { int trials = minutesToTest / minutesToDie + 1; int pigs = 0; while (pow(trials, pigs) <= buckets) { pigs++; } return pigs; } int main() { int buckets = 4, minutesToDie = 15, minutesToTest = 15; cout << poorPigs(buckets, minutesToDie, minutesToTest) << endl; }

#include <bits/stdc++.h> using namespace std; int poorPigs(int buckets, int minutesToDie, int minutesToTest) { int trials = minutesToTest / minutesToDie + 1; int pigs = 0; while (pow(trials - 1, pigs) < buckets) { pigs++; } return pigs; } int main() { int buckets = 4, minutesToDie = 15, minutesToTest = 15; cout << poorPigs(buckets, minutesToDie, minutesToTest) << endl; }

You are given an array prices where prices[i] is the price of a given stock on the i th day. Find the maximum profit you can achieve. You may complete at most two transactions . Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again). Example 1: Input: prices = [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3. Example 2: Input: prices = [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again. Example 3: Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0. Constraints: 1 <= prices.length <= 10 5 0 <= prices[i] <= 10 5

int maxProfit(vector<int>& prices) { int n = prices.size(); if(n == 0) return 0; vector<int> left(n, 0), right(n, 0); int minPrice = prices[0]; for(int i = 1; i < n; i++) { minPrice = min(minPrice, prices[i]); left[i] = max(left[i-1], prices[i] - minPrice); } int maxPrice = prices[n-1]; for(int i = n-2; i >= 0; i--) { maxPrice = max(maxPrice, prices[i]); right[i] = max(right[i+1], maxPrice - prices[i]); } int maxProfit = 0; for(int i = 0; i < n; i++) { maxProfit = max(maxProfit, left[i] + right[i]); } return maxProfit; }

int maxProfit(vector<int>& prices) { int n = prices.size(); if(n == 0) return 0; vector<int> left(n, 0), right(n, 0); int minPrice = prices[0]; for(int i = 1; i < n; i++) { minPrice = min(minPrice, prices[i]); left[i] = max(left[i-1], prices[i] - minPrice); } int maxPrice = prices[n-1]; for(int i = n-2; i >= 0; i--) { maxPrice = min(maxPrice, prices[i]); right[i] = max(right[i+1], maxPrice - prices[i]); } int maxProfit = 0; for(int i = 0; i < n; i++) { maxProfit = max(maxProfit, left[i] + right[i]); } return maxProfit; }

int maxProfit(vector<int>& prices) { int n = prices.size(); if(n == 0) return 0; vector<int> left(n, 0), right(n, 0); int minPrice = prices[0]; for(int i = 1; i < n; i++) { minPrice = min(minPrice, prices[i]); left[i] = max(left[i-1], prices[i] - minPrice); } int maxPrice = prices[n-1]; for(int i = n-2; i >= 0; i--) { maxPrice = max(maxPrice, prices[i]); right[i] = max(right[i+1], maxPrice - prices[i]); } int maxProfit = 0; for(int i = 0; i < n; i++) { maxProfit = max(maxProfit, left[i] + left[i]); } return maxProfit; }

int maxProfit(vector<int>& prices) { int n = prices.size(); if(n == 0) return 0; vector<int> left(n, 0), right(n, 0); int minPrice = INT_MAX; for(int i = 1; i < n; i++) { minPrice = min(minPrice, prices[i]); left[i] = max(left[i-1], prices[i] - minPrice); } int maxPrice = prices[n-1]; for(int i = n-2; i >= 0; i--) { maxPrice = max(maxPrice, prices[i]); right[i] = max(right[i+1], maxPrice - prices[i]); } int maxProfit = 0; for(int i = 0; i < n; i++) { maxProfit = max(maxProfit, left[i] + right[i]); } return maxProfit; }

An array is squareful if the sum of every pair of adjacent elements is a perfect square . Given an integer array nums, return the number of permutations of nums that are squareful . Two permutations perm1 and perm2 are different if there is some index i such that perm1[i] != perm2[i]. Example 1: Input: nums = [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: nums = [2,2,2] Output: 1 Constraints: 1 <= nums.length <= 12 0 <= nums[i] <= 10 9

#include <bits/stdc++.h> using namespace std; bool isSquare(int x) { int r = sqrt(x); return r * r == x; } int dfs(vector<int>& nums, vector<int>& used, int prev, int count) { if (count == nums.size()) return 1; int res = 0; for (int i = 0; i < nums.size(); i++) { if (used[i]) continue; if (i > 0 && nums[i] == nums[i-1] && !used[i-1]) continue; if (prev == -1 || isSquare(prev + nums[i])) { used[i] = 1; res += dfs(nums, used, nums[i], count + 1); used[i] = 0; } } return res; } int numSquarefulPerms(vector<int>& nums) { sort(nums.begin(), nums.end()); vector<int> used(nums.size(), 0); return dfs(nums, used, -1, 0); } int main() { vector<int> nums = {1,17,8}; cout << numSquarefulPerms(nums); }

#include <bits/stdc++.h> using namespace std; bool isSquare(int x) { int r = sqrt(x); return r * r == x; } int dfs(vector<int>& nums, vector<int>& used, int prev, int count) { if (count == nums.size()) return 1; int res = 0; for (int i = 0; i < nums.size(); i++) { if (used[i]) continue; if (i > 0 && nums[i] == nums[i-1] && !used[i-1]) continue; if (prev == -1 || isSquare(prev + nums[i])) { used[i] = 1; res += dfs(nums, used, nums[i], count + 1); used[i] = 1; } } return res; } int numSquarefulPerms(vector<int>& nums) { sort(nums.begin(), nums.end()); vector<int> used(nums.size(), 0); return dfs(nums, used, -1, 0); } int main() { vector<int> nums = {1,17,8}; cout << numSquarefulPerms(nums); }

#include <bits/stdc++.h> using namespace std; bool isSquare(int x) { int r = sqrt(x); return r * r == x; } int dfs(vector<int>& nums, vector<int>& used, int prev, int count) { if (count == nums.size()) return 1; int res = 0; for (int i = 0; i < nums.size(); i++) { if (used[i]) continue; if (i > 0 && nums[i] == nums[i-1] && !used[i-1]) continue; if (prev == -1 || !isSquare(prev + nums[i])) { used[i] = 1; res += dfs(nums, used, nums[i], count + 1); used[i] = 0; } } return res; } int numSquarefulPerms(vector<int>& nums) { sort(nums.begin(), nums.end()); vector<int> used(nums.size(), 0); return dfs(nums, used, -1, 0); } int main() { vector<int> nums = {1,17,8}; cout << numSquarefulPerms(nums); }

#include <bits/stdc++.h> using namespace std; bool isSquare(int x) { int r = sqrt(x); return r * r == x; } int dfs(vector<int>& nums, vector<int>& used, int prev, int count) { if (count == nums.size()) return 1; int res = 0; for (int i = 0; i < nums.size(); i++) { if (used[i]) continue; if (i > 0 && nums[i] == nums[i-1] && !used[i-1]) continue; if (prev == -1 || isSquare(abs(prev - nums[i]))) { used[i] = 1; res += dfs(nums, used, nums[i], count + 1); used[i] = 0; } } return res; } int numSquarefulPerms(vector<int>& nums) { sort(nums.begin(), nums.end()); vector<int> used(nums.size(), 0); return dfs(nums, used, -1, 0); } int main() { vector<int> nums = {1,17,8}; cout << numSquarefulPerms(nums); }

You are given an array arr which consists of only zeros and ones, divide the array into three non-empty parts such that all of these parts represent the same binary value. If it is possible, return any [i, j] with i + 1 < j, such that: arr[0], arr[1], ..., arr[i] is the first part, arr[i + 1], arr[i + 2], ..., arr[j - 1] is the second part, and arr[j], arr[j + 1], ..., arr[arr.length - 1] is the third part. All three parts have equal binary values. If it is not possible, return [-1, -1]. Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed , so [0,1,1] and [1,1] represent the same value. Example 1: Input: arr = [1,0,1,0,1] Output: [0,3] Example 2: Input: arr = [1,1,0,1,1] Output: [-1,-1] Example 3: Input: arr = [1,1,0,0,1] Output: [0,2] Constraints: 3 <= arr.length <= 3 * 10 4 arr[i] is 0 or 1

#include <bits/stdc++.h> using namespace std; vector<int> threeEqualParts(vector<int>& arr) { int n = arr.size(); int ones = 0; for (int x : arr) ones += x; if (ones == 0) return {0, n - 1}; if (ones % 3 != 0) return {-1, -1}; int k = ones / 3; int first = -1, second = -1, third = -1, cnt = 0; for (int i = 0; i < n; i++) { if (arr[i] == 1) { cnt++; if (cnt == 1) first = i; else if (cnt == k + 1) second = i; else if (cnt == 2 * k + 1) third = i; } } while (third < n && arr[first] == arr[second] && arr[third] == arr[first]) { first++; second++; third++; } if (third == n) return {first - 1, second}; return {-1, -1}; }

#include <bits/stdc++.h> using namespace std; vector<int> threeEqualParts(vector<int>& arr) { int n = arr.size(); int ones = 0; for (int x : arr) ones += x; if (ones == 0) return {0, n - 1}; if (ones % 3 != 0) return {-1, -1}; int k = ones / 3; int first = -1, second = -1, third = -1, cnt = 0; for (int i = 0; i < n; i++) { if (arr[i] == 1) { cnt++; if (cnt == 1) first = i; else if (cnt == k) second = i; else if (cnt == 2 * k) third = i; } } while (third < n && arr[first] == arr[second] && arr[third] == arr[first]) { first++; second++; third++; } if (third == n) return {first - 1, second}; return {-1, -1}; }

#include <bits/stdc++.h> using namespace std; vector<int> threeEqualParts(vector<int>& arr) { int n = arr.size(); int ones = 0; for (int x : arr) ones += x; if (ones == 0) return {0, n - 1}; if (ones % 3 != 0) return {-1, -1}; int k = ones / 3; int first = -1, second = -1, third = -1, cnt = 0; for (int i = 0; i < n; i++) { if (arr[i] == 1) { cnt++; if (cnt == 1) first = i; else if (cnt == k + 1) second = i; else if (cnt == 2 * k + 1) third = i; } } while (third < n && arr[first] == arr[second] && arr[third] == arr[first]) { first++; second++; third++; } if (second == n) return {first - 1, second}; return {-1, -1}; }

#include <bits/stdc++.h> using namespace std; vector<int> threeEqualParts(vector<int>& arr) { int n = arr.size(); int ones = 0; for (int x : arr) ones += x; if (ones == 0) return {0, n - 1}; if (ones % 3 != 0) return {-1, -1}; int k = ones / 3; int first = -1, second = -1, third = -1, cnt = 0; for (int i = 0; i < n; i++) { if (arr[i] == 1) { cnt++; if (cnt == 1) first = i; else if (cnt == k + 1) second = i; else if (cnt == 2 * k + 1) third = i; } } while (third < n && arr[first] == arr[second] && arr[third] == arr[first]) { first++; second++; third++; } if (third == n) return {first, second}; return {-1, -1}; }

Given an integer array nums, return the number of AND triples . An AND triple is a triple of indices (i, j, k) such that: 0 <= i < nums.length 0 <= j < nums.length 0 <= k < nums.length nums[i] & nums[j] & nums[k] == 0, where & represents the bitwise-AND operator. Example 1: Input: nums = [2,1,3] Output: 12 Explanation: We could choose the following i, j, k triples: (i=0, j=0, k=1) : 2 & 2 & 1 (i=0, j=1, k=0) : 2 & 1 & 2 (i=0, j=1, k=1) : 2 & 1 & 1 (i=0, j=1, k=2) : 2 & 1 & 3 (i=0, j=2, k=1) : 2 & 3 & 1 (i=1, j=0, k=0) : 1 & 2 & 2 (i=1, j=0, k=1) : 1 & 2 & 1 (i=1, j=0, k=2) : 1 & 2 & 3 (i=1, j=1, k=0) : 1 & 1 & 2 (i=1, j=2, k=0) : 1 & 3 & 2 (i=2, j=0, k=1) : 3 & 2 & 1 (i=2, j=1, k=0) : 3 & 1 & 2 Example 2: Input: nums = [0,0,0] Output: 27 Constraints: 1 <= nums.length <= 1000 0 <= nums[i] < 2 16

#include <bits/stdc++.h> using namespace std; int main() { vector<int> nums = {2,1,3}; int n = nums.size(); int ans = 0; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ int andVal = nums[i] & nums[j]; for(int k=0;k<n;k++){ if((andVal & nums[k]) == 0) ans++; } } } cout << ans; }

#include <bits/stdc++.h> using namespace std; int main() { vector<int> nums = {2,1,3}; int n = nums.size(); int ans = 0; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ for(int k=0;k<n;k++){ if((nums[i] & nums[j] & nums[k]) == nums[i]) ans++; } } } cout << ans; }

#include <bits/stdc++.h> using namespace std; int main() { vector<int> nums = {2,1,3}; int n = nums.size(); int ans = 0; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ int andVal = nums[i] & nums[j]; for(int k=0;k<n;k++){ if((andVal & nums[k]) != 1) ans++; } } } cout << ans; }

#include <bits/stdc++.h> using namespace std; int main() { vector<int> nums = {2,1,3}; int n = nums.size(); int ans = 0; for(int i=0;i<n;i++){ for(int j=i;j<n;j++){ int andVal = nums[i] & nums[j]; for(int k=0;k<n;k++){ if((andVal & nums[k]) == 0) ans++; } } } cout << ans; }

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KIỂU DỮ LIỆU DANH SÁCH

Professional Development

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53 Qs

Mock 2027 Batch

Quiz

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Computers

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Professional Development

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Practice Problem

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Hard

Anshul Oza

Used 17+ times

FREE Resource

FILL IN THE BLANK QUESTION

1 min • 1 pt

Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 am
and travels towards B at 20 kmph. Another train starts from B at 8 am and
travels towards A at a speed of 25 kmph. At what time will they meet?(Directly Mention the result without comma or unit)

FILL IN THE BLANK QUESTION

1 min • 1 pt

The average age of 8 persons is increased by 2 years, when two of them,
whose ages are 20 years and 24 years, are replaced by two women. What is
the average age of these women?(Directly Mention the result without comma or unit)

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Mock 2027 Batch

54 questions

Enter Your Name

Enter Your Enrolment

Enter Your Branch(CSE, CSE-DS, CSE-AIML, CY, CSIT, IT, EC, EC-VLSI)

Enter Your Acropolis Email ID(xxxxxxxxx@acropolis.in)

A man bought a bike at the marked price of Rs.65,000 with a discount of 3% on its
marked price. If the salesman makes 30% of profit on this sale then what will be the cost
price of the bike?(Directly mention the answer without comma or unit)

Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 am
and travels towards B at 20 kmph. Another train starts from B at 8 am and
travels towards A at a speed of 25 kmph. At what time will they meet?(Directly Mention the result without comma or unit)

The average age of 8 persons is increased by 2 years, when two of them,
whose ages are 20 years and 24 years, are replaced by two women. What is
the average age of these women?(Directly Mention the result without comma or unit)

How much water be added to 14 litres of milk worth Rs. 5.40 a litre so that the
value of the mixture is Rs. 4.20 a litre?(Directly Mention the result without comma or unit)

Two pipes can fill a tank in 25 and 30 minutes respectively and a waste pipe
can empty 5 gallons per minute. All the three pipes working together can fill
the tank in 15 minutes. The capacity of the tank is in gallons?(Directly Mention the result without comma or unit)

Create a free account and access millions of resources

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Computers

Mock 2027 Batch

54 questions

Enter Your Name

Enter Your Enrolment

Enter Your Branch(CSE, CSE-DS, CSE-AIML, CY, CSIT, IT, EC, EC-VLSI)

Enter Your Acropolis Email ID(xxxxxxxxx@acropolis.in)

A man bought a bike at the marked price of Rs.65,000 with a discount of 3% on itsmarked price. If the salesman makes 30% of profit on this sale then what will be the costprice of the bike?(Directly mention the answer without comma or unit)

Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 amand travels towards B at 20 kmph. Another train starts from B at 8 am andtravels towards A at a speed of 25 kmph. At what time will they meet?(Directly Mention the result without comma or unit)

The average age of 8 persons is increased by 2 years, when two of them,whose ages are 20 years and 24 years, are replaced by two women. What isthe average age of these women?(Directly Mention the result without comma or unit)

How much water be added to 14 litres of milk worth Rs. 5.40 a litre so that thevalue of the mixture is Rs. 4.20 a litre?(Directly Mention the result without comma or unit)

Two pipes can fill a tank in 25 and 30 minutes respectively and a waste pipecan empty 5 gallons per minute. All the three pipes working together can fillthe tank in 15 minutes. The capacity of the tank is in gallons?(Directly Mention the result without comma or unit)

Create a free account and access millions of resources

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Computers

A man bought a bike at the marked price of Rs.65,000 with a discount of 3% on its
marked price. If the salesman makes 30% of profit on this sale then what will be the cost
price of the bike?(Directly mention the answer without comma or unit)

Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 am
and travels towards B at 20 kmph. Another train starts from B at 8 am and
travels towards A at a speed of 25 kmph. At what time will they meet?(Directly Mention the result without comma or unit)

The average age of 8 persons is increased by 2 years, when two of them,
whose ages are 20 years and 24 years, are replaced by two women. What is
the average age of these women?(Directly Mention the result without comma or unit)

How much water be added to 14 litres of milk worth Rs. 5.40 a litre so that the
value of the mixture is Rs. 4.20 a litre?(Directly Mention the result without comma or unit)

Two pipes can fill a tank in 25 and 30 minutes respectively and a waste pipe
can empty 5 gallons per minute. All the three pipes working together can fill
the tank in 15 minutes. The capacity of the tank is in gallons?(Directly Mention the result without comma or unit)