Friday, September 16, 2016

Maximize number of 0s by flipping a subarray - GeeksforGeeks


Maximize number of 0s by flipping a subarray - GeeksforGeeks
Given a binary array, find the maximum number zeros in an array with one flip of a subarray allowed. A flip operation switches all 0s to 1s and 1s to 0s.

This problem can be reduced to largest subarray sum problem. The idea is to consider every 0 as -1 and every 1 as 1, find the sum of largest subarray sum in this modified array. This sum is our required max_diff ( count of 0s – count of 1s in any subarray). Finally we return the max_diff plus count of zeros in original array.
// A Kadane's algorithm based solution to find maximum
// number of 0s by flipping a subarray.
int findMaxZeroCount(bool arr[], int n)
{
    // Initialize count of zeros and maximum difference
    // between count of 1s and 0s in a subarray
    int orig_zero_count = 0;
 
    // Initiale overall max diff for any subarray
    int max_diff = 0;
 
    // Initialize current diff
    int curr_max = 0;
 
    for (int i=0; i<n; i++)
    {
        // Count of zeros in original array (Not related
        // to Kadane's algorithm)
        if (arr[i] == 0)
           orig_zero_count++;
 
        // Value to be considered for finding maximum sum
        int val = (arr[i] == 1)? 1 : -1;
 
        // Update current max and max_diff
        curr_max = max(val, curr_max + val);
        max_diff = max(max_diff, curr_max);
    }
    max_diff = max(0, max_diff);
 
    return orig_zero_count + max_diff;
}

A simple solution is to consider all subarrays and find a subarray with maximum value of (count of 1s) – (count of 0s). Let this value be max_diff. Finally return count of zeros in original array plus max_diff.
// A Kadane's algorithm based solution to find maximum
// number of 0s by flipping a subarray.
int findMaxZeroCount(bool arr[], int n)
{
    // Initialize max_diff = maximum of (Count of 0s -
    // count of 1s) for all subarrays.
    int max_diff = 0;
 
    // Initialize count of 0s in original array
    int orig_zero_count = 0;
 
    // Consider all Subarrays by using two nested two
    // loops
    for (int i=0; i<n; i++)
    {
        // Increment count of zeros
        if (arr[i] == 0)
            orig_zero_count++;
 
        // Initialize counts of 0s and 1s
        int count1 = 0, count0 = 0;
 
        // Consider all subarrays starting from arr[i]
        // and find the difference between 1s and 0s.
        // Update max_diff if required
        for (int j=i; j<n; j++)
        {
            (arr[j] == 1)? count1++ : count0++;
            max_diff = max(max_diff, count1 - count0);
        }
    }
 
    // Final result would be count of 0s in original
    // array plus max_diff.
    return orig_zero_count + max_diff;
}
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