Saturday, December 26, 2015

Longest Span with same Sum in two Binary arrays - GeeksforGeeks


Longest Span with same Sum in two Binary arrays - GeeksforGeeks
Given two binary arrays arr1[] and arr2[] of same size n. Find length of the longest common span (i, j) where j >= i such that arr1[i] + arr1[i+1] + …. + arr1[j] = arr2[i] + arr2[i+1] + …. + arr2[j].
Input: arr1[] = {0, 1, 0, 0, 0, 0};
       arr2[] = {1, 0, 1, 0, 0, 1};
Output: 4
The longest span with same sun is from index 1 to 4.
Method 2 (Using Auxiliary Array)
Time Complexity: Θ(n)
Auxiliary Space: Θ(n)
The idea is based on below observations.
  1. Since there are total n elements, maximum sum is n for both arrays.
  2. Difference between two sums varies from -n to n. So there are total 2n + 1 possible values of difference.
  3. If differences between prefix sums of two arrays become same at two points, then subarrays between these two points have same sum.
Below is Complete Algorithm.
  1. Create an auxiliary array of size 2n+1 to store starting points of all possible values of differences (Note that possible values of differences vary from -n to n, i.e., there are total 2n+1 possible values)
  2. Initialize starting points of all differences as -1.
  3. Initialize maxLen as 0 and prefix sums of both arrays as 0, preSum1 = 0, preSum2 = 0
  4. Travers both arrays from i = 0 to n-1.
    1. Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]
    2. Compute difference of current prefix sums: curr_diff = preSum1 – preSum2
    3. Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -n
    4. If curr_diff is 0, then i+1 is maxLen so far
    5. Else If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as i
    6. Else (curr_diff is NOT seen first time), then consider i as ending point and find length of current same sum span. If this length is more, then update maxLen
  5. Return maxLen
int longestCommonSum(bool arr1[], bool arr2[], int n)
{
    // Initialize result
    int maxLen = 0;
    // Initialize prefix sums of two arrays
    int preSum1 = 0, preSum2 = 0;
    // Create an array to store staring and ending
    // indexes of all possible diff values. diff[i]
    // would store starting and ending points for
    // difference "i-n"
    int diff[2*n+1];
    // Initialize all starting and ending values as -1.
    memset(diff, -1, sizeof(diff));
    // Traverse both arrays
    for (int i=0; i<n; i++)
    {
        // Update prefix sums
        preSum1 += arr1[i];
        preSum2 += arr2[i];
        // Comput current diff and index to be used
        // in diff array. Note that diff can be negative
        // and can have minimum value as -1.
        int curr_diff = preSum1 - preSum2;
        int diffIndex = n + curr_diff;
        // If current diff is 0, then there are same number
        // of 1's so far in both arrays, i.e., (i+1) is
        // maximum length.
        if (curr_diff == 0)
            maxLen = i+1;
        // If current diff is seen first time, then update
        // starting index of diff.
        else if ( diff[diffIndex] == -1)
            diff[diffIndex] = i;
        // Current diff is already seen
        else
        {
            // Find lenght of this same sum common span
            int len = i - diff[diffIndex];
            // Update max len if needed
            if (len > maxLen)
                maxLen = len;
        }
    }
    return maxLen;
}

Method 1 (Simple Solution)
One by one by consider same subarrays of both arrays. For all subarrays, compute sums and if sums are same and current length is more than max length, then update max length. 
int longestCommonSum(bool arr1[], bool arr2[], int n)
{
    // Initialize result
    int maxLen = 0;
    // One by one pick all possible starting points
    // of subarrays
    for (int i=0; i<n; i++)
    {
       // Initialize sums of current subarrays
       int sum1 = 0, sum2 = 0;
       // Conider all points for starting with arr[i]
       for (int j=i; j<n; j++)
       {
           // Update sums
           sum1 += arr1[j];
           sum2 += arr2[j];
           // If sums are same and current length is
           // more than maxLen, update maxLen
           if (sum1 == sum2)
           {
             int len = j-i+1;
             if (len > maxLen)
                maxLen = len;
           }
       }
    }
    return maxLen;
}


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