Given a 2D square matrix arr[][] of dimensions N x N, the task is to find the maximum path sum by moving diagonally from any cell and each cell must be visited only once i.e., from the cell (i, j), a player can move to the cell (i + 1, j + 1).
Examples:
Input: arr[][] = {{1, 2, 3}, {3, 5, 10}, {1 3 5}}
Output: 12
Explanation:
Sum of cells (1, 1), (2, 2) and (3, 3) is 11.
The sum of cells (1, 2), (2, 3) and (1, 3) is 3.
The sum of cells (2, 1) and (3, 2) is 6.
The sum of cell (3, 1) is 1.
The maximum possible sum is 12.Input: arr[][] = {{1, 1, 1}, {1 1 1}, {1 1 1}}
Output: 3
Approach: To solve this problem, the idea is to traverse the matrix diagonally for first row and column elements and sum up their diagonal elements within the range of the matrix.
Follow the steps below to solve the problem:
- Initialize a variable, say max with 0.
- Choose each cell (i, j) from the first row and from the first column.
- Now, from each cell, find the diagonal sum starting from that cell by incrementing i and j by 1, say sum.
- Then, update max as max(max, sum).
- After traversing, print max as the required answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum sum int MaximumSum(vector<vector< int > >& arr, int n)
{ int ans = 0;
// Loop to traverse through the
// upper triangular matrix and
// update the maximum sum to ans
for ( int i = 0; i < n; i++) {
int x = 0, y = i, sum = 0;
for ( int j = i; j < n; j++) {
sum += arr[x++][y++];
}
if (sum > ans)
ans = sum;
}
// Traverse through the
// lower triangular matrix
for ( int i = 1; i < n; i++) {
int x = i, y = 0, sum = 0;
for ( int j = i; j < n; j++) {
sum += arr[x++][y++];
}
if (sum > ans)
ans = sum;
}
return ans;
} // Driver Code int main()
{ // Given matrix
vector<vector< int > > arr;
arr = { { 1, 2, 3 },
{ 3, 5, 10 },
{ 1, 3, 5 } };
// Given dimension
int n = arr.size();
cout << MaximumSum(arr, n);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the maximum sum static int MaximumSum( int [][]arr, int n)
{ int ans = 0 ;
// Loop to traverse through the
// upper triangular matrix and
// update the maximum sum to ans
for ( int i = 0 ; i < n; i++)
{
int x = 0 , y = i, sum = 0 ;
for ( int j = i; j < n; j++)
{
sum += arr[x++][y++];
}
if (sum > ans)
ans = sum;
}
// Traverse through the
// lower triangular matrix
for ( int i = 1 ; i < n; i++)
{
int x = i, y = 0 , sum = 0 ;
for ( int j = i; j < n; j++)
{
sum += arr[x++][y++];
}
if (sum > ans)
ans = sum;
}
return ans;
} // Driver Code public static void main(String[] args)
{ // Given matrix
int [][]arr = { { 1 , 2 , 3 },
{ 3 , 5 , 10 },
{ 1 , 3 , 5 } };
// Given dimension
int n = arr.length;
System.out.print(MaximumSum(arr, n));
} } // This code is contributed by 29AjayKumar |
# Python3 program for the above approach # Function to find the maximum sum def MaximumSum(arr, n):
ans = 0 ;
# Loop to traverse through the
# upper triangular matrix and
# update the maximum sum to ans
for i in range (n):
x, y, sum = 0 , i, 0
for j in range (i, n):
sum , x, y = sum + arr[x][y], x + 1 , y + 1
if ( sum > ans):
ans = sum
# Traverse through the
# lower triangular matrix
for i in range ( 1 , n):
x, y, sum = i, 0 , 0
for j in range (i, n):
sum , x, y = sum + arr[x][y], x + 1 , y + 1
if ( sum > ans):
ans = sum
return ans
# Driver Code if __name__ = = '__main__' :
# Given matrix
arr = [ [ 1 , 2 , 3 ],
[ 3 , 5 , 10 ],
[ 1 , 3 , 5 ]]
# Given dimension
n = len (arr)
print (MaximumSum(arr, n))
# This code is contributed by mohit kumar 29
|
// C# program for the above approach using System;
class GFG{
// Function to find the maximum sum
static int MaximumSum( int [,]arr, int n)
{
int ans = 0;
// Loop to traverse through the
// upper triangular matrix and
// update the maximum sum to ans
for ( int i = 0; i < n; i++)
{
int x = 0, y = i, sum = 0;
for ( int j = i; j < n; j++)
{
sum += arr[x++, y++];
}
if (sum > ans)
ans = sum;
}
// Traverse through the
// lower triangular matrix
for ( int i = 1; i < n; i++)
{
int x = i, y = 0, sum = 0;
for ( int j = i; j < n; j++)
{
sum += arr[x++, y++];
}
if (sum > ans)
ans = sum;
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
// Given matrix
int [,]arr = { { 1, 2, 3 },
{ 3, 5, 10 },
{ 1, 3, 5 } };
// Given dimension
int n = arr.GetLength(0);
Console.Write(MaximumSum(arr, n));
}
} // This code is contributed by shikhasingrajput |
<script> // Javascript program of the above approach // Function to find the maximum sum function MaximumSum(arr, n)
{ let ans = 0;
// Loop to traverse through the
// upper triangular matrix and
// update the maximum sum to ans
for (let i = 0; i < n; i++)
{
let x = 0, y = i, sum = 0;
for (let j = i; j < n; j++)
{
sum += arr[x++][y++];
}
if (sum > ans)
ans = sum;
}
// Traverse through the
// lower triangular matrix
for (let i = 1; i < n; i++)
{
let x = i, y = 0, sum = 0;
for (let j = i; j < n; j++)
{
sum += arr[x++][y++];
}
if (sum > ans)
ans = sum;
}
return ans;
} // Driver Code
// Given matrix
let arr = [[ 1, 2, 3 ],
[ 3, 5, 10 ],
[ 1, 3, 5 ]];
// Given dimension
let n = arr.length;
document.write(MaximumSum(arr, n));
</script> |
12
Time Complexity: O(N2)
Auxiliary Space: O(1)