1004. Max Consecutive Ones III

[Problem]

Array Binary Search Sliding Window Prefix Sum

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Intuition

• Interpret the question as: “Longest subarray with at most k zeroes”.
• There is NO ACTUAL FLIPPING you need to do.
• The question is reduced to finding the maxImum subarray length with at most k zeroes.

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Approach 1. Sliding Window

• Use a sliding window defined by left and right pointers.
• Increment the right pointer while counting zeroes in the window.
• If the count of zeroes exceeds k, increment the left pointer to reduce the window size until the zero count is valid.
• Update the maximum window size during the process.

Complexity Analysis

• Time Complexity: O(n)
  • Each element is processed at most twice (once by right and once by left).
• Space Complexity: O(1)
  • No extra space used.

class Solution {
    public int longestOnes(int[] nums, int k) {
        int n = nums.length;
        
        // start left from index `0`, and initialize maxLen to `0`
        // `maxLen` is the length of longest window where the no. of zeroes <= k
        int left = 0, maxLen = 0;

        // iterate with the right pointer to "expand" the window
        for (int right = 0; right < n; right++) {
            // if a `0` is encountered, decrement `k`, which tracks the no. of zeroes allowed in the window
            if (nums[right] == 0) {
                k--;
            }

            // if the no. of zeroes in the window exceeds `k`, move the `left` pointer to shrink the window
            // This ensures we always maintain at most `k` zeroes in the window
            while (k < 0) {
                // if the element at the `left` pointer is 0, increment `k` to "restore" a zero
                if (nums[left] == 0) {
                    k++;
                }
                // move the `left` pointer forward to shrink the window
                left++;
            }

            // update `maxLen` with the maximum window length encountered so far
            maxLen = Math.max(maxLen, right - left + 1);
        }
        return maxLen;
    }
}

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Approach 2. Prefix Sum

Because the array contains only 0s and 1s, the sum in a given range [left, right] represents the count of 1s in that window.

• For example, if the sum is 4, it means there are 4 ones in that window, and the rest are zeros.

• The number of zeroes can be calculated as:
  No. of zeroes = window_length - No. of ones

• Now, to solve the problem:
  • Increment the right pointer to expand the window, allowing up to k zeroes in the window. Continue expanding until there are at most k zeroes.
  • If the number of zeroes exceeds k, start incrementing the left pointer.
  • Incrementing left effectively shifts the window. Initially, only right was expanding the window length, but by moving left, the window size becomes constant while adjusting the zero count.
  • Continue this process until the number of zeroes in the window becomes less than or equal to k.

Complexity Analysis

• Time Complexity: O(n)
  One pass to calculate the prefix sum, and one pass to adjust the window.
• Space Complexity: O(n)
  Additional space for the prefix sum array.

Code

class Solution {
    public int longestOnes(int[] nums, int k) {
        int n = nums.length;
        
        // initialize the prefixSum array to store cumulative sums
        // prefixSum[i] will represent the sum of elements in nums[0] to nums[i-1]
        int[] prefixSum = new int[n + 1];

        // compute the prefix sum array
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        // start left from index `0`, and initialize maxLen to `0`
        // `maxLen` is the length of longest window where the no. of zeroes <= k
        int left = 0, maxLen = 0;

        // iterate with the right pointer to "expand" the window
        for (int right = 0; right < n; right++) {
            // calculate the no. of 1s in the current window [left, right]
            int numOnes = prefixSum[right + 1] - prefixSum[left];

            // calculate the no. of 0s in the current window
            // remember, `no. of zeros = window_length - no. of ones`
            int numZeroes = (right - left + 1) - numOnes;

            // if the no. of zeroes exceeds k, increment `left`
            // this will shrink the window to make the no. of zeroes <= k
            if (numZeroes > k) {
                left++;
            } else {
                // if the window is valid, update `maxLen`
                maxLen = Math.max(maxLen, right - left + 1);
            }
        }
        return maxLen;
    }
}