Given an array of 0 and 1, find minimum no. of swaps to bring all 1s together (only adjacent swaps allowed)

Question

If given an array of 1\'s and 0\'s, what\'s good algorithm to show the minimum number of adjacent swaps needed to group all of the 1\'s together. The 1\'s don\'t need to be

Answer 1

Approach : This can be done by finding number of zeroes to the right side of every 1 and add them. In order to sort the array every one always has to perform a swap operation with every zero on its right side.

So the total number of swap operations for a particular 1 in array is the number of zeroes on its right hand side. Find the number of zeroes on right side for every one i.e. the number of swaps and add them all to obtain the total number of swaps.

// Java code to find minimum number of swaps to sort a binary array

class MinimumNumberOfSwapsNeeded { 

    static int findMinSwaps(int arr[], int n) 
    { 
        // Array to store count of zeroes 
        int noOfZeroes[] = new int[n]; 
        int i, count = 0; 

        // Count number of zeroes 
        // on right side of every one. 
        noOfZeroes[n - 1] = 1 - arr[n - 1]; 
        for (i = n - 2; i >= 0; i--)  
        { 
            noOfZeroes[i] = noOfZeroes[i + 1]; 
            if (arr[i] == 0) 
                noOfZeroes[i]++; 
        } 

        // Count total number of swaps by adding number 
        // of zeroes on right side of every one. 
        for (i = 0; i < n; i++)  
        { 
            if (arr[i] == 1) 
                count += noOfZeroes[i]; 
        } 
        return count; 
    }       
    // Driver Code 
    public static void main(String args[]) 
    { 
        int ar[] = { 0, 0, 1, 0, 1, 0, 1, 1 }; 
        System.out.println(findMinSwaps(ar, ar.length)); 
    } 
}