Minimum sum submatrix in a given 2D array

Given a 2D array, find the minimum sum submatrix in it.

Examples:  

Input : M[][] = {{1, 2, -1, -4, -20},
                 {-8, -3, 4, 2, 1},
                 {3, 8, 10, 1, 3},
                 {-4, -1, 1, 7, -6}}
Output : -26
Submatrix starting from (Top, Left): (0, 0)
and ending at (Bottom, Right): (1, 4) indexes.
The elements are of the submatrix are:
{ {1, 2, -1, -4, -20},
  {-8, -3, 4, 2, 1}  } having sum = -26

Method 1 (Naive Approach): Check every possible submatrix in a given 2D array. This solution requires 4 nested loops and the time complexity of this solution would be O(n^4).

Method 2 (Efficient Approach):

Kadane’s algorithm for the 1D array can be used to reduce the time complexity to O(n^3). The idea is to fix the left and right columns one by one and find the minimum sum contiguous rows for every left and right column pair. We basically find top and bottom row numbers (which have minimum sum) for every fixed left and right column pair. To find the top and bottom row numbers, calculate the sum of elements in every row from left to right and store these sums in an array say temp[]. So temp[i] indicates the sum of elements from left to right in row i. 

If we apply Kadane’s 1D algorithm on temp[] and get the minimum sum subarray of temp, this minimum sum would be the minimum possible sum with left and right as boundary columns. To get the overall minimum sum, we compare this sum with the minimum sum so far.

Implementation:

C++

// C++ implementation to find minimum sum
// submatrix in a given 2D array
#include <bits/stdc++.h>
using namespace std;

#define ROW 4
#define COL 5

// Implementation of Kadane's algorithm for
// 1D array. The function returns the minimum
// sum and stores starting and ending indexes
// of the minimum sum subarray at addresses
// pointed by start and finish pointers
// respectively.
int kadane(int* arr, int* start, int* finish,
                                       int n)
{
    // initialize sum, maxSum and
    int sum = 0, minSum = INT_MAX, i;

    // Just some initial value to check for
    // all negative values case
    *finish = -1;

    // local variable
    int local_start = 0;

    for (i = 0; i < n; ++i) {
        sum += arr[i];
        if (sum > 0) {
            sum = 0;
            local_start = i + 1;
        } else if (sum < minSum) {
            minSum = sum;
            *start = local_start;
            *finish = i;
        }
    }

    // There is at-least one non-negative number
    if (*finish != -1)
        return minSum;

    // Special Case: When all numbers in arr[]
    // are positive
    minSum = arr[0];
    *start = *finish = 0;

    // Find the minimum element in array
    for (i = 1; i < n; i++) {
        if (arr[i] < minSum) {
            minSum = arr[i];
            *start = *finish = i;
        }
    }
    return minSum;
}

// function to find minimum sum submatrix
// in a given 2D array
void findMinSumSubmatrix(int M[][COL])
{
    // Variables to store the final output
    int minSum = INT_MAX, finalLeft, finalRight,
                          finalTop, finalBottom;

    int left, right, i;
    int temp[ROW], sum, start, finish;

    // Set the left column
    for (left = 0; left < COL; ++left) {

        // Initialize all elements of temp as 0
        memset(temp, 0, sizeof(temp));

        // Set the right column for the left
        // column set by outer loop
        for (right = left; right < COL; ++right) {

            // Calculate sum between current left
            // and right for every row 'i'
            for (i = 0; i < ROW; ++i)
                temp[i] += M[i][right];

            // Find the minimum sum subarray in temp[].
            // The kadane() function also sets values
            // of start and finish.  So 'sum' is sum of
            // rectangle between (start, left) and
            // (finish, right) which is the minimum sum
            // with boundary columns strictly as
            // left and right.
            sum = kadane(temp, &start, &finish, ROW);

            // Compare sum with maximum sum so far. If
            // sum is more, then update maxSum and other
            // output values
            if (sum < minSum) {
                minSum = sum;
                finalLeft = left;
                finalRight = right;
                finalTop = start;
                finalBottom = finish;
            }
        }
    }

    // Print final values
    cout << "(Top, Left): (" << finalTop << ", "
            << finalLeft << ")\n";
    cout << "(Bottom, Right): (" << finalBottom << ", "
         << finalRight << ")\n";
    cout << "Minimum sum: " << minSum;
}

// Driver program to test above
int main()
{
    int M[ROW][COL] = { { 1, 2, -1, -4, -20 },
                        { -8, -3, 4, 2, 1 },
                        { 3, 8, 10, 1, 3 },
                        { -4, -1, 1, 7, -6 } };
    findMinSumSubmatrix(M);
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;

class GFG
{
    static int ROW = 4;
    static int COL = 5;
    static int start;
    static int finish;

    static int kadane(int[] arr, int n)
    {

        // initialize sum, maxSum and
        int sum = 0, minSum = Integer.MAX_VALUE, i;

        // Just some initial value to check for
        // all negative values case
        finish = -1;

        // local variable
        int local_start = 0;

        for (i = 0; i < n; ++i)
        {
            sum += arr[i];
            if (sum > 0)
            {
                sum = 0;
                local_start = i + 1;
            }
          else if (sum < minSum)
          {
                minSum = sum;
                start = local_start;
                finish = i;
            }
        }

        // There is at-least one non-negative number
        if (finish != -1)
            return minSum;

        // Special Case: When all numbers in arr[]
        // are positive
        minSum = arr[0];
        start = finish = 0;

        // Find the minimum element in array
        for (i = 1; i < n; i++)
        {
            if (arr[i] < minSum)
            {
                minSum = arr[i];
                start = finish = i;
            }
        }
        return minSum;

    }

    // function to find minimum sum submatrix
    // in a given 2D array
    static void findMinSumSubmatrix(int[][] M)
    {

        // Variables to store the final output
        int minSum = Integer.MAX_VALUE;
        int finalLeft = 0 , finalRight = 0, finalTop = 0, finalBottom = 0;
        int left, right, i;
        int []temp= new int[ROW];
        int sum;

        // Set the left column
        for (left = 0; left < COL; ++left)
        {

            // Initialize all elements of temp as 0
            Arrays.fill(temp, 0);

            // Set the right column for the left
            // column set by outer loop
            for (right = left; right < COL; ++right)
            {

                // Calculate sum between current left
                // and right for every row 'i'
                for (i = 0; i < ROW; ++i)
                    temp[i] += M[i][right];

                // Find the minimum sum subarray in temp[].
                // The kadane() function also sets values
                // of start and finish.  So 'sum' is sum of
                // rectangle between (start, left) and
                // (finish, right) which is the minimum sum
                // with boundary columns strictly as
                // left and right.
                sum = kadane(temp, ROW);

                // Compare sum with maximum sum so far. If
                // sum is more, then update maxSum and other
                // output values
                if (sum < minSum)
                {
                    minSum = sum;
                    finalLeft = left;
                    finalRight = right;
                    finalTop = start;
                    finalBottom = finish;
                }
            }

        }

        // Print final values
        System.out.println("(Top, Left): (" +
                           finalTop + ", " +
                           finalLeft + ")");
        System.out.println("(Bottom, Right): (" +
                           finalBottom + ", " +
                           finalRight + ")");
        System.out.println("Minimum sum: "+ minSum);

    }

    // Driver program to test above
    public static void main (String[] args)
    {
        int[][] M ={{ 1, 2, -1, -4, -20 },
                    { -8, -3, 4, 2, 1 },
                    { 3, 8, 10, 1, 3 },
                    { -4, -1, 1, 7, -6 }};
        findMinSumSubmatrix(M);
    }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation to find minimum
# sum submatrix in a given 2D array
import sys

# Implementation of Kadane's algorithm for
# 1D array. The function returns the minimum
# sum and stores starting and ending indexes
# of the minimum sum subarray at addresses
# pointed by start and finish pointers
# respectively.
def kadane(arr, start, finish, n):

    # Initialize sum, maxSum and
    sum, minSum = 0, sys.maxsize

    # Just some initial value to check
    # for all negative values case
    finish = -1

    # Local variable
    local_start = 0

    for i in range(n):
        sum += arr[i]

        if (sum > 0):
            sum = 0
            local_start = i + 1

        elif (sum < minSum):
            minSum = sum
            start = local_start
            finish = i

    # There is at-least one non-negative number
    if (finish != -1):
        return minSum, start, finish

    # Special Case: When all numbers in arr[]
    # are positive
    minSum = arr[0]
    start, finish = 0, 0

    # Find the minimum element in array
    for i in range(1, n):
        if (arr[i] < minSum):
            minSum = arr[i]
            start = finish = i

    return minSum, start, finish

# Function to find minimum sum submatrix
# in a given 2D array
def findMinSumSubmatrix(M):

    # Variables to store the final output
    minSum = sys.maxsize
    finalLeft = 0
    finalRight = 0
    finalTop = 0
    finalBottom = 0

    #left, right, i = 0, 0, 0
    sum, start, finish = 0, 0, 0

    # Set the left column
    for left in range(5):

        # Initialize all elements of temp as 0
        temp = [0 for i in range(5)]

        # Set the right column for the left
        # column set by outer loop
        for right in range(left, 5):

            # Calculate sum between current left
            # and right for every row 'i'
            for i in range(4):
                temp[i] += M[i][right]

            # Find the minimum sum subarray in temp[].
            # The kadane() function also sets values
            # of start and finish. So 'sum' is sum of
            # rectangle between (start, left) and
            # (finish, right) which is the minimum sum
            # with boundary columns strictly as
            # left and right.
            sum, start, finish = kadane(temp, start,
                                        finish, 4)

            # Compare sum with maximum sum so far. If
            # sum is more, then update maxSum and other
            # output values
            if (sum < minSum):
                minSum = sum
                finalLeft = left
                finalRight = right
                finalTop = start
                finalBottom = finish

    # Print final values
    print("(Top, Left): (", finalTop,
          ",", finalLeft, ")")
    print("(Bottom, Right): (", finalBottom,
          ",", finalRight, ")")
    print("Minimum sum:", minSum)

# Driver code
if __name__ == '__main__':

    M = [ [ 1, 2, -1, -4, -20 ],
          [ -8, -3, 4, 2, 1 ],
          [ 3, 8, 10, 1, 3 ],
          [ -4, -1, 1, 7, -6 ] ]

    findMinSumSubmatrix(M)

# This code is contributed by mohit kumar 29

C#

using System;
public class GFG
{
    static int ROW = 4;
    static int COL = 5;
    static int start;
    static int finish;
    static int kadane(int[] arr, int n)
    {

        // initialize sum, maxSum and
        int sum = 0, minSum = Int32.MaxValue, i;

        // Just some initial value to check for
        // all negative values case
        finish = -1;

        // local variable
        int local_start = 0;

        for (i = 0; i < n; ++i)
        {
            sum += arr[i];
            if (sum > 0)
            {
                sum = 0;
                local_start = i + 1;
            }
          else if (sum < minSum)
          {
                minSum = sum;
                start = local_start;
                finish = i;
            }
        }

        // There is at-least one non-negative number
        if (finish != -1)
            return minSum;

        // Special Case: When all numbers in arr[]
        // are positive
        minSum = arr[0];
        start = finish = 0;

        // Find the minimum element in array
        for (i = 1; i < n; i++)
        {
            if (arr[i] < minSum)
            {
                minSum = arr[i];
                start = finish = i;
            }
        }
        return minSum;
    }

    // function to find minimum sum submatrix
    // in a given 2D array
    static void findMinSumSubmatrix(int[,] M)
    {

        // Variables to store the final output
        int minSum = Int32.MaxValue;
        int finalLeft = 0 , finalRight = 0, finalTop = 0, finalBottom = 0;
        int left, right, i;
        int []temp= new int[ROW];
        int sum;

        // Set the left column
        for (left = 0; left < COL; ++left)
        {

            // Initialize all elements of temp as 0
            Array.Fill(temp, 0);

            // Set the right column for the left
            // column set by outer loop
            for (right = left; right < COL; ++right)
            {

                // Calculate sum between current left
                // and right for every row 'i'
                for (i = 0; i < ROW; ++i)
                    temp[i] += M[i, right];

                // Find the minimum sum subarray in temp[].
                // The kadane() function also sets values
                // of start and finish.  So 'sum' is sum of
                // rectangle between (start, left) and
                // (finish, right) which is the minimum sum
                // with boundary columns strictly as
                // left and right.
                sum = kadane(temp, ROW);

                // Compare sum with maximum sum so far. If
                // sum is more, then update maxSum and other
                // output values
                if (sum < minSum)
                {
                    minSum = sum;
                    finalLeft = left;
                    finalRight = right;
                    finalTop = start;
                    finalBottom = finish;
                }
            }

        }
        Console.WriteLine("(Top, Left): (" +
                          finalTop + ", " +
                          finalLeft + ")");
        Console.WriteLine("(Bottom, Right): (" +
                          finalBottom + ", " +
                          finalRight + ")");
        Console.WriteLine("Minimum sum: "+ minSum);
    }

    // Driver program to test above
    static public void Main ()
    {
        int[,] M ={{ 1, 2, -1, -4, -20 },
                   { -8, -3, 4, 2, 1 },
                   { 3, 8, 10, 1, 3 },
                   { -4, -1, 1, 7, -6 }};
        findMinSumSubmatrix(M);
    }
}

// This code is contributed by rag2127

Javascript

<script>

/*package whatever //do not write
package name here */

    var ROW = 4;
    var COL = 5;
    var start;
    var finish;

    function kadane(arr , n) {

        // initialize sum, maxSum and
        var sum = 0, minSum = Number.MAX_VALUE, i;

        // Just some initial value to check for
        // all negative values case
        finish = -1;

        // local variable
        var local_start = 0;

        for (i = 0; i < n; ++i) {
            sum += arr[i];
            if (sum > 0) {
                sum = 0;
                local_start = i + 1;
            } else if (sum < minSum) {
                minSum = sum;
                start = local_start;
                finish = i;
            }
        }

        // There is at-least one non-negative number
        if (finish != -1)
            return minSum;

        // Special Case: When all numbers in arr
        // are positive
        minSum = arr[0];
        start = finish = 0;

        // Find the minimum element in array
        for (i = 1; i < n; i++) {
            if (arr[i] < minSum) {
                minSum = arr[i];
                start = finish = i;
            }
        }
        return minSum;

    }

    // function to find minimum sum submatrix
    // in a given 2D array
    function findMinSumSubmatrix(M)
    {

        // Variables to store the final output
        var minSum = Number.MAX_VALUE;
        var finalLeft = 0, finalRight = 0, finalTop = 0,
        finalBottom = 0;
        var left, right, i;
        var temp = Array(ROW).fill(0);
        var sum;

        // Set the left column
        for (left = 0; left < COL; ++left) {

            // Initialize all elements of temp as 0
            temp.fill(0);

            // Set the right column for the left
            // column set by outer loop
            for (right = left; right < COL; ++right)
            {

                // Calculate sum between current left
                // and right for every row 'i'
                for (i = 0; i < ROW; ++i)
                    temp[i] += M[i][right];

                // Find the minimum sum subarray in temp.
                // The kadane() function also sets values
                // of start and finish. So 'sum' is sum of
                // rectangle between (start, left) and
                // (finish, right) which is the minimum sum
                // with boundary columns strictly as
                // left and right.
                sum = kadane(temp, ROW);

                // Compare sum with maximum sum so far. If
                // sum is more, then update maxSum and other
                // output values
                if (sum < minSum) {
                    minSum = sum;
                    finalLeft = left;
                    finalRight = right;
                    finalTop = start;
                    finalBottom = finish;
                }
            }

        }

        // Print final values
        document.write("(Top, Left): (" + finalTop + ", "
        + finalLeft + ")<br/>");
        document.write("(Bottom, Right): (" + finalBottom + ", "
        + finalRight + ")<br/>");
        document.write("Minimum sum: " + minSum);

    }

    // Driver program to test above

        var M = [ [ 1, 2, -1, -4, -20 ],
                [ -8, -3, 4, 2, 1 ],
                [ 3, 8, 10, 1, 3 ],
                [ -4, -1, 1, 7, -6 ] ];
        findMinSumSubmatrix(M);

// This code contributed by umadevi9616

</script>

Output

(Top, Left): (0, 0)
(Bottom, Right): (1, 4)
Minimum sum: -26

Time Complexity: O(n³)
Auxiliary Space: O(ROW)

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