1004

1004.java

class Solution {
    public int longestOnes(int[] nums, int k) {
        
        int n = nums.length;
        
        int countOfZero = 0;
        
        int longest = 0;
        //.   s       
        // [0,1,1,0,0,0,1,1,1,1,0]
        // sliding window template
        int s = 0;
        for (int f = 0; f < n; f++) {
            // check the element af num[f]
            if (nums[f] == 0) {
                countOfZero++;
            }
            
            // we got more 0s than allowed ?
            while (countOfZero > k) {
                // start move slow pointer until we have k zeros between nums[s..f]
                if (nums[s] == 0) countOfZero--;
                s++;
            }
            
            // we got a candidate
            longest = Math.max(longest, f-s+1);
            
            /*
            
            // when reached threshold 
            while (countOfZero == k) { // 1,1,0,0,1,1,1,0,1,1
                
                
                for (int i = s; i <= f; i++) System.out.print(nums[i] + ",");
                // System.out.println("-----");
                
                
                // between s and f, we have a potential candidate
                longest = Math.max(longest, f-s+1);
                
                // continue move the slow pointer until the threshold is broken
                if (nums[s] == 0) countOfZero--;
                s++;
            }
            
            // countOfZero < k
            // System.out.println("longest: " + longest);
            
            */
            
        }
        
        return longest;   
    }
    //                                            s.                    f
    // [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1] // 1,1,0,0,1,1,1,0,1,1   0
    //. 3
    
}