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Maximum sum path in a matrix from top to bottom and back

  • Difficulty Level : Hard
  • Last Updated : 15 Jul, 2021

Given a matrix of dimension N * M. The task is find the maximum sum of path from arr[0][0] to arr[N – 1][M – 1] and back from arr[N – 1][M – 1] to arr[0][0]
On the path from arr[0][0] to arr[N – 1][M – 1], you can traverse in down and right directions and on the path from arr[N – 1][M – 1] to arr[0][0], you can traverse in up and left directions. 
Note: Both the path must not be equal i.e. there has to be at least one cell arr[i][j] which is not common in both the paths.
Examples: 
 

Input: 
mat[][]= {{1, 0, 3, -1},
          {3, 5, 1, -2},
          {-2, 0, 1, 1},
          {2, 1, -1, 1}}
Output: 16

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Maximum sum on path from arr[0][0] to arr[3][3] 
= 1 + 3 + 5 + 1 + 1 + 1 + 1 = 13
Maximum sum on path from arr[3][3] to arr[0][0] = 3
Total path sum = 13 + 3 = 16

Input: 
mat[][]= {{1, 0},
          {1, 1}}
Output: 3

Approach: This problem is somewhat similar to Minimum Cost Path problem except that in the present problem, two paths with maximum sum are to be found. Also, we need to take care that cells on both paths contribute only once to the sum. 
First thing to notice is that path from arr[N – 1][M – 1] to arr[0][0] is nothing but another path from arr[0][0] to arr[N – 1][M – 1]. So, we have to find two paths from arr[0][0] to arr[N – 1][M – 1] with maximum sum. 
Approaching in a similar way as Minimum Cost Path problem, we start both paths from arr[0][0] together and recur to neighbouring cells of the matrix till we reach arr[N – 1][M – 1]. To make sure that a cell doesn’t contribute more than once, we check if current cell on both path are the same or not. If they are same, it is added to the answer only once.
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Input matrix
int n = 4, m = 4;
int arr[4][4] = { { 1, 0, 3, -1 },
                  { 3, 5, 1, -2 },
                  { -2, 0, 1, 1 },
                  { 2, 1, -1, 1 } };
 
// DP matrix
int cache[5][5][5];
 
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
int sum(int i1, int j1, int i2, int j2)
{
    if (i1 == i2 && j1 == j2) {
        return arr[i1][j1];
    }
    return arr[i1][j1] + arr[i2][j2];
}
 
// Recursive function to return the
// required maximum cost path
int maxSumPath(int i1, int j1, int i2)
{
 
    // Column number of second path
    int j2 = i1 + j1 - i2;
 
    // Base Case
    if (i1 >= n || i2 >= n || j1 >= m || j2 >= m) {
        return 0;
    }
 
    // If already calculated, return from DP matrix
    if (cache[i1][j1][i2] != -1) {
        return cache[i1][j1][i2];
    }
    int ans = INT_MIN;
 
    // Recurring for neighbouring cells of both paths together
    ans = max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));
    ans = max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));
 
    // Saving result to the DP matrix for current state
    cache[i1][j1][i2] = ans;
 
    return ans;
}
 
// Driver code
int main()
{
    memset(cache, -1, sizeof(cache));
    cout << maxSumPath(0, 0, 0);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Input matrix
static int n = 4, m = 4;
static int arr[][] = { { 1, 0, 3, -1 },
                { 3, 5, 1, -2 },
                { -2, 0, 1, 1 },
                { 2, 1, -1, 1 } };
 
// DP matrix
static int cache[][][] = new int[5][5][5];
 
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
static int sum(int i1, int j1, int i2, int j2)
{
    if (i1 == i2 && j1 == j2)
    {
        return arr[i1][j1];
    }
    return arr[i1][j1] + arr[i2][j2];
}
 
// Recursive function to return the
// required maximum cost path
static int maxSumPath(int i1, int j1, int i2)
{
 
    // Column number of second path
    int j2 = i1 + j1 - i2;
 
    // Base Case
    if (i1 >= n || i2 >= n || j1 >= m || j2 >= m)
    {
        return 0;
    }
 
    // If already calculated, return from DP matrix
    if (cache[i1][j1][i2] != -1)
    {
        return cache[i1][j1][i2];
    }
    int ans = Integer.MIN_VALUE;
 
    // Recurring for neighbouring cells of both paths together
    ans = Math.max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = Math.max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));
    ans = Math.max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = Math.max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));
 
    // Saving result to the DP matrix for current state
    cache[i1][j1][i2] = ans;
 
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    //set initial value
    for(int i=0;i<5;i++)
    for(int i1=0;i1<5;i1++)
    for(int i2=0;i2<5;i2++)
    cache[i][i1][i2]=-1;
     
    System.out.println( maxSumPath(0, 0, 0));
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python 3 implementation of the approach
import sys
 
# Input matrix
n = 4
m = 4
arr = [[1, 0, 3, -1],
    [3, 5, 1, -2],
    [-2, 0, 1, 1],
    [2, 1, -1, 1]]
 
# DP matrix
cache = [[[-1 for i in range(5)] for j in range(5)] for k in range(5)]
 
# Function to return the sum of the cells
# arr[i1][j1] and arr[i2][j2]
def sum(i1, j1, i2, j2):
    if (i1 == i2 and j1 == j2):
        return arr[i1][j1]
    return arr[i1][j1] + arr[i2][j2]
 
# Recursive function to return the
# required maximum cost path
def maxSumPath(i1, j1, i2):
     
    # Column number of second path
    j2 = i1 + j1 - i2
 
    # Base Case
    if (i1 >= n or i2 >= n or j1 >= m or j2 >= m):
        return 0
 
    # If already calculated, return from DP matrix
    if (cache[i1][j1][i2] != -1):
        return cache[i1][j1][i2]
    ans = -sys.maxsize-1
 
    # Recurring for neighbouring cells of both paths together
    ans = max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2))
    ans = max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2))
    ans = max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2))
    ans = max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2))
 
    # Saving result to the DP matrix for current state
    cache[i1][j1][i2] = ans
 
    return ans
 
# Driver code
if __name__ == '__main__':
    print(maxSumPath(0, 0, 0))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Input matrix
static int n = 4, m = 4;
static int [,]arr = { { 1, 0, 3, -1 },
                { 3, 5, 1, -2 },
                { -2, 0, 1, 1 },
                { 2, 1, -1, 1 } };
 
// DP matrix
static int [,,]cache = new int[5, 5, 5];
 
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
static int sum(int i1, int j1, int i2, int j2)
{
    if (i1 == i2 && j1 == j2)
    {
        return arr[i1, j1];
    }
    return arr[i1, j1] + arr[i2, j2];
}
 
// Recursive function to return the
// required maximum cost path
static int maxSumPath(int i1, int j1, int i2)
{
 
    // Column number of second path
    int j2 = i1 + j1 - i2;
 
    // Base Case
    if (i1 >= n || i2 >= n || j1 >= m || j2 >= m)
    {
        return 0;
    }
 
    // If already calculated, return from DP matrix
    if (cache[i1, j1, i2] != -1)
    {
        return cache[i1, j1, i2];
    }
    int ans = int.MinValue;
 
    // Recurring for neighbouring cells of both paths together
    ans = Math.Max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = Math.Max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));
    ans = Math.Max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = Math.Max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));
 
    // Saving result to the DP matrix for current state
    cache[i1, j1, i2] = ans;
 
    return ans;
}
 
// Driver code
public static void Main(String []args)
{
    //set initial value
    for(int i = 0; i < 5; i++)
        for(int i1 = 0; i1 < 5; i1++)
            for(int i2 = 0; i2 < 5; i2++)
                cache[i,i1,i2]=-1;
     
    Console.WriteLine( maxSumPath(0, 0, 0));
}
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
// Javascript implementation of the approach
 
// Input matrix
let n = 4, m = 4;
let arr=[[ 1, 0, 3, -1 ],
                [ 3, 5, 1, -2 ],
                [ -2, 0, 1, 1 ],
                [ 2, 1, -1, 1 ]];
                 
// DP matrix
let cache=new Array(5);
for(let i = 0; i < 5; i++)
{
    cache[i] = new Array(5);
    for(let j = 0; j < 5; j++)
    {
        cache[i][j] = new Array(5);
        for(let k = 0; k < 5; k++)
        {
            cache[i][j][k] = -1;
        }
    }
}
 
// Function to return the sum of the cells
// arr[i1][j1] and arr[i2][j2]
function sum(i1, j1, i2, j2)
{
    if (i1 == i2 && j1 == j2)
    {
        return arr[i1][j1];
    }
    return arr[i1][j1] + arr[i2][j2];
}
 
// Recursive function to return the
// required maximum cost path
function maxSumPath(i1,j1,i2)
{
    // Column number of second path
    let j2 = i1 + j1 - i2;
   
    // Base Case
    if (i1 >= n || i2 >= n || j1 >= m || j2 >= m)
    {
        return 0;
    }
   
    // If already calculated, return from DP matrix
    if (cache[i1][j1][i2] != -1)
    {
        return cache[i1][j1][i2];
    }
    let ans = Number.MIN_VALUE;
   
    // Recurring for neighbouring cells of both paths together
    ans = Math.max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = Math.max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));
    ans = Math.max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));
    ans = Math.max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));
   
    // Saving result to the DP matrix for current state
    cache[i1][j1][i2] = ans;
   
    return ans;
}
 
// Driver code
document.write( maxSumPath(0, 0, 0));
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
16

 

Time Complexity: O((N2) * M)
 




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