Swapping Apples or Oranges

Problem : Requirement is that you are given a list of randomly arranged apples & oranges. You will be given some number of
                chances, say N to replace the oranges with apples. Now you have to find the maximum number of continuous
                apples you can have after those many number of swaps. To keep the problem straight & bit simple, we can replace
                the orange with apple from outside the list.
Solution : Below code is not the optimal one & I feel that it can be improved further but currently putting this solution here.
                  I have represented apple with 1 & orange with 0.

public class Interview {
 public static void main(String[] args) {
  int[] fruits = {1,1,1,0,1,0,1,0,0,1,1,1,1,1,0,0,0,0,1,1,1,1,1,1,0};
  int swaps = 4;
  int maxContOnes = 0;
  int k = 0;
  int swap = swaps;
  int start = 0;
  for(int i : fruits) {
   if(i == 0) {
    if(swap > 0)
     swap--;
    // There is no chance to swap any zero so move the start counter to next value.
    else if(swaps == 0) {
     start = k+1;
    }
    else {
     // Set the max count value till the current pointer as all available swaps are consumed now.
     if(maxContOnes < k-start) {
      maxContOnes = k - start;
     }
     // if swap = 0 then move the start pointer till one zero is crossed.
     while(swap == 0 && start < k) {
      if(fruits[start] == 0)
       swap++;
      start++;
     }
     // Minus the count of zero by one to count the current zero.
     swap--;
    }
   }
   k++;
  }
  // This is to calculate the value once more if any zero was processed in the last of above iteration.
  if(maxContOnes < k-start) {
   maxContOnes = k - start;
  }
  System.out.println(maxContOnes);
 }
}

One more way to implement for the above problem in a concise way is shown below as contributed by Jatin Agrawal -

​public class FindConsecutiveApplesSeqMaxLength {
 public static void main(String[] args) {
//  System.out.println(getSeqMaxLength(new char[]{'A','A','O','O','A','O','A'}, 3));
//  System.out.println(getSeqMaxLength(new char[]{'A','A','O','O','A','O','A'}, 2));
  long start = System.nanoTime();
  System.out.println(getSeqMaxLength(new char[] {'A','A','A','O','A','O','A','O','O','A','A','A','A','A','O','O','O','O','A','A','A','A','A','A','O'}, 4));
  System.out.println(System.nanoTime() - start);
 }
 private static int getSeqMaxLength(char[] seq, int k) {
  int seqMaxLength = 0;
  int remaining = k;
  int curSeqStart = -1;
  int i = 0;
  for (; i < seq.length; i++) {
   if (seq[i] == 'O') {
    if (remaining == 0) {
     int curSeqLength = i - 1 - curSeqStart;
     if (curSeqLength > seqMaxLength) {
      seqMaxLength = curSeqLength;
     }
     while (seq[++curSeqStart] != 'O');
    } else {
     remaining--;
    }
   }
  }
  int curSeqLength = i - 1 - curSeqStart;
  if (curSeqLength > seqMaxLength) {
   seqMaxLength = curSeqLength;
  }
  return seqMaxLength;
 }
}