maxConsecutiveOne.md

1004. Max Consecutive Ones III, Medium, 2 pointers

The atmost method

Approach

• since we can switch up to k elements therefore we will prepare a sliding window of k size.
• meaning, we will trigger the internal while loop only when count_of_zero > k

code

int longestOnes(vector<int>& nums, int k) {
    int count = 0;
    int ans = 0;
    int l = 0;
    for (int r = 0; r < nums.size(); r++) {
      count += nums[r] == 0;
      while (count > k and l <= r) {
        count -= nums[l] == 0;
        l++;
      }
      ans = max(ans, r - l + 1);
    }
    return ans;
  }