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Codeforces|Leetcode|Codechef free solutions

Codeforces|Leetcode|Codechef free solutions

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Free codeforces, Codechef, Leetcode solutions are available 😍😍😍😍😍😍 Helped More than 200+ students to crack coding round in 2022 and helped placed them in Good companies. 🥳🥳🥳🤩🤩🤩 Dm @Cpsoln if you want help in coding round.

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Leetcode C 😊😁

    int numberOfAlternatingGroups(vector& colors, int k) {         int n = colors.size();         if (n < k) return 0;         int cntrr = 0;         bool isAlternating = true;         for (int i = 0; i < k - 1; ++i) {             if (colors[i] == colors[i + 1]) {                 isAlternating = false;                 break;             }         }         if (isAlternating) cntrr++;         for (int i = 1; i < n; ++i) {             int outgoing = colors[(i - 1) % n];             int incoming = colors[(i + k - 1) % n];             int next = colors[(i + k) % n];             int prev = colors[(i - 2 + n) % n];             if (outgoing == prev) isAlternating = false;             if (incoming == next) isAlternating = false;             if (outgoing != prev && incoming != next) isAlternating = true;             if (isAlternating) cntrr++;         }         return cntrr;     }

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long long countSubarrays(vector& nums, int k) {         long long count = 0;         int n = nums.size();         for (int start = 0; start < n; ++start) {             int current_and = nums[start];             if (current_and == k) {                 count++;             }             for (int end = start + 1; end < n; ++end) {                 current_and &= nums[end];                 if (current_and == k) {                     count++;                 } else if (current_and < k) {                     break; // If current_and < k, no need to extend further                 }             }         }         return count;     }

int countAlternatingGroups(vector& colors) { int n = colors.size(); int count = 0; for (int i = 0; i < n; ++i) { if (colors[i] != colors[(i+1) % n] && colors[i] == colors[(i+2) % n]) { count++; } } return count; }