Find a specific pair in Matrix

Given an n x n matrix mat[n][n] of integers, find the maximum value of mat(c, d) – mat(a, b) over all choices of indexes such that both c > a and d > b.

Example:

Input:
mat[N][N] = {{ 1, 2, -1, -4, -20 },
             { -8, -3, 4, 2, 1 }, 
             { 3, 8, 6, 1, 3 },
             { -4, -1, 1, 7, -6 },
             { 0, -4, 10, -5, 1 }};
Output: 18
The maximum value is 18 as mat[4][2] 
- mat[1][0] = 18 has maximum difference.

The program should do only ONE traversal of the matrix. i.e. expected time complexity is O(n²)

A simple solution would be to apply Brute-Force. For all values mat(a, b) in the matrix, we find mat(c, d) that has maximum value such that c > a and d > b and keeps on updating maximum value found so far. We finally return the maximum value.

Below is its implementation.

C++

// A Naive method to find maximum value of mat[d][e] 
// - ma[a][b] such that d > a and e > b 
#include <bits/stdc++.h> 
using namespace std; 
#define N 5 
  
// The function returns maximum value A(d,e) - A(a,b) 
// over all choices of indexes such that both d > a 
// and e > b. 
int findMaxValue(int mat[][N]) 
{ 
    // stores maximum value 
    int maxValue = INT_MIN; 
  
    // Consider all possible pairs mat[a][b] and 
    // mat[d][e] 
    for (int a = 0; a < N - 1; a++) 
    for (int b = 0; b < N - 1; b++) 
        for (int d = a + 1; d < N; d++) 
        for (int e = b + 1; e < N; e++) 
            if (maxValue < (mat[d][e] - mat[a][b])) 
                maxValue = mat[d][e] - mat[a][b]; 
  
    return maxValue; 
} 
  
// Driver program to test above function 
int main() 
{ 
int mat[N][N] = { 
                { 1, 2, -1, -4, -20 }, 
                { -8, -3, 4, 2, 1 }, 
                { 3, 8, 6, 1, 3 }, 
                { -4, -1, 1, 7, -6 }, 
                { 0, -4, 10, -5, 1 } 
            }; 
    cout << "Maximum Value is "
        << findMaxValue(mat); 
  
    return 0; 
} 

Java

// A Naive method to find maximum value of mat1[d][e] 
// - ma[a][b] such that d > a and e > b 
import java.io.*; 
import java.util.*; 
   
class GFG  
{ 
    // The function returns maximum value A(d,e) - A(a,b) 
    // over all choices of indexes such that both d > a 
    // and e > b. 
    static int findMaxValue(int N,int mat[][]) 
    { 
        // stores maximum value 
        int maxValue = Integer.MIN_VALUE; 
       
        // Consider all possible pairs mat[a][b] and 
        // mat1[d][e] 
        for (int a = 0; a < N - 1; a++) 
          for (int b = 0; b < N - 1; b++) 
             for (int d = a + 1; d < N; d++) 
               for (int e = b + 1; e < N; e++) 
                  if (maxValue < (mat[d][e] - mat[a][b])) 
                      maxValue = mat[d][e] - mat[a][b]; 
       
        return maxValue; 
    } 
       
    // Driver code 
    public static void main (String[] args)  
    { 
        int N = 5; 
  
        int mat[][] = { 
                      { 1, 2, -1, -4, -20 }, 
                      { -8, -3, 4, 2, 1 }, 
                      { 3, 8, 6, 1, 3 }, 
                      { -4, -1, 1, 7, -6 }, 
                      { 0, -4, 10, -5, 1 } 
                   }; 
  
        System.out.print("Maximum Value is " +  
                         findMaxValue(N,mat)); 
    } 
} 
   
// This code is contributed 
// by Prakriti Gupta 

Python 3

# A Naive method to find maximum  
# value of mat[d][e] - mat[a][b] 
# such that d > a and e > b 
N = 5
  
# The function returns maximum  
# value A(d,e) - A(a,b) over  
# all choices of indexes such  
# that both d > a and e > b. 
def findMaxValue(mat): 
      
    # stores maximum value 
    maxValue = 0
  
    # Consider all possible pairs  
    # mat[a][b] and mat[d][e] 
    for a in range(N - 1): 
        for b in range(N - 1): 
            for d in range(a + 1, N): 
                for e in range(b + 1, N): 
                    if maxValue < int (mat[d][e] - 
                                       mat[a][b]): 
                        maxValue = int(mat[d][e] - 
                                       mat[a][b]); 
  
    return maxValue; 
  
# Driver Code 
mat = [[ 1, 2, -1, -4, -20 ], 
       [ -8, -3, 4, 2, 1 ], 
       [ 3, 8, 6, 1, 3 ], 
       [ -4, -1, 1, 7, -6 ], 
       [ 0, -4, 10, -5, 1 ]]; 
         
print("Maximum Value is " + 
       str(findMaxValue(mat))) 
        
# This code is contributed  
# by ChitraNayal 

C#

// A Naive method to find maximum  
// value of mat[d][e] - mat[a][b] 
// such that d > a and e > b 
using System; 
class GFG 
{ 
      
    // The function returns 
    // maximum value A(d,e) - A(a,b) 
    // over all choices of indexes  
    // such that both d > a 
    // and e > b. 
    static int findMaxValue(int N,  
                            int [,]mat) 
    { 
          
        //stores maximum value 
        int maxValue = int.MinValue; 
      
        // Consider all possible pairs  
        // mat[a][b] and mat[d][e] 
        for (int a = 0; a< N - 1; a++) 
        for (int b = 0; b < N - 1; b++) 
            for (int d = a + 1; d < N; d++) 
            for (int e = b + 1; e < N; e++) 
                if (maxValue < (mat[d, e] -  
                                mat[a, b])) 
                    maxValue = mat[d, e] -  
                               mat[a, b]; 
  
        return maxValue; 
    } 
      
    // Driver code 
    public static void Main ()  
    { 
        int N = 5; 
  
        int [,]mat = {{1, 2, -1, -4, -20}, 
                      {-8, -3, 4, 2, 1}, 
                      {3, 8, 6, 1, 3}, 
                      {-4, -1, 1, 7, -6}, 
                      {0, -4, 10, -5, 1}}; 
        Console.Write("Maximum Value is " +  
                      findMaxValue(N,mat)); 
    } 
} 
  
// This code is contributed  
// by ChitraNayal 

PHP

<?php 
// A Naive method to find maximum  
// value of $mat[d][e] - ma[a][b] 
// such that $d > $a and $e > $b 
$N = 5; 
  
// The function returns maximum  
// value A(d,e) - A(a,b) over  
// all choices of indexes such  
// that both $d > $a and $e > $b. 
function findMaxValue(&$mat) 
{ 
    global $N; 
      
    // stores maximum value 
    $maxValue = PHP_INT_MIN; 
  
    // Consider all possible  
    // pairs $mat[$a][$b] and 
    // $mat[$d][$e] 
    for ($a = 0; $a < $N - 1; $a++) 
    for ($b = 0; $b < $N - 1; $b++) 
        for ($d = $a + 1; $d < $N; $d++) 
        for ($e = $b + 1; $e < $N; $e++) 
            if ($maxValue < ($mat[$d][$e] -  
                             $mat[$a][$b])) 
                $maxValue = $mat[$d][$e] -  
                            $mat[$a][$b]; 
  
    return $maxValue; 
} 
  
// Driver Code 
$mat = array(array(1, 2, -1, -4, -20), 
             array(-8, -3, 4, 2, 1), 
             array(3, 8, 6, 1, 3), 
             array(-4, -1, 1, 7, -6), 
             array(0, -4, 10, -5, 1)); 
              
echo "Maximum Value is " .  
       findMaxValue($mat); 
  
// This code is contributed  
// by ChitraNayal 
?> 

Output:

Maximum Value is 18

The above program runs in O(n^4) time which is nowhere close to expected time complexity of O(n^2)

An efficient solution uses extra space. We pre-process the matrix such that index(i, j) stores max of elements in matrix from (i, j) to (N-1, N-1) and in the process keeps on updating maximum value found so far. We finally return the maximum value.

C++

// An efficient method to find maximum value of mat[d] 
// - ma[a][b] such that c > a and d > b 
#include <bits/stdc++.h> 
using namespace std; 
#define N 5 
  
// The function returns maximum value A(c,d) - A(a,b) 
// over all choices of indexes such that both c > a 
// and d > b. 
int findMaxValue(int mat[][N]) 
{ 
    //stores maximum value 
    int maxValue = INT_MIN; 
  
    // maxArr[i][j] stores max of elements in matrix 
    // from (i, j) to (N-1, N-1) 
    int maxArr[N][N]; 
  
    // last element of maxArr will be same's as of 
    // the input matrix 
    maxArr[N-1][N-1] = mat[N-1][N-1]; 
  
    // preprocess last row 
    int maxv = mat[N-1][N-1];  // Initialize max 
    for (int j = N - 2; j >= 0; j--) 
    { 
        if (mat[N-1][j] > maxv) 
            maxv = mat[N - 1][j]; 
        maxArr[N-1][j] = maxv; 
    } 
  
    // preprocess last column 
    maxv = mat[N - 1][N - 1];  // Initialize max 
    for (int i = N - 2; i >= 0; i--) 
    { 
        if (mat[i][N - 1] > maxv) 
            maxv = mat[i][N - 1]; 
        maxArr[i][N - 1] = maxv; 
    } 
  
    // preprocess rest of the matrix from bottom 
    for (int i = N-2; i >= 0; i--) 
    { 
        for (int j = N-2; j >= 0; j--) 
        { 
            // Update maxValue 
            if (maxArr[i+1][j+1] - mat[i][j] > 
                                            maxValue) 
                maxValue = maxArr[i + 1][j + 1] - mat[i][j]; 
  
            // set maxArr (i, j) 
            maxArr[i][j] = max(mat[i][j], 
                               max(maxArr[i][j + 1], 
                                   maxArr[i + 1][j]) ); 
        } 
    } 
  
    return maxValue; 
} 
  
// Driver program to test above function 
int main() 
{ 
    int mat[N][N] = { 
                      { 1, 2, -1, -4, -20 }, 
                      { -8, -3, 4, 2, 1 }, 
                      { 3, 8, 6, 1, 3 }, 
                      { -4, -1, 1, 7, -6 }, 
                      { 0, -4, 10, -5, 1 } 
                    }; 
    cout << "Maximum Value is " 
         << findMaxValue(mat); 
  
    return 0; 
} 

Java

// An efficient method to find maximum value of mat1[d] 
// - ma[a][b] such that c > a and d > b 
import java.io.*; 
import java.util.*; 
   
class GFG  
{ 
    // The function returns maximum value A(c,d) - A(a,b) 
    // over all choices of indexes such that both c > a 
    // and d > b. 
    static int findMaxValue(int N,int mat[][]) 
    { 
        //stores maximum value 
        int maxValue = Integer.MIN_VALUE; 
       
        // maxArr[i][j] stores max of elements in matrix 
        // from (i, j) to (N-1, N-1) 
        int maxArr[][] = new int[N][N]; 
       
        // last element of maxArr will be same's as of 
        // the input matrix 
        maxArr[N-1][N-1] = mat[N-1][N-1]; 
       
        // preprocess last row 
        int maxv = mat[N-1][N-1];  // Initialize max 
        for (int j = N - 2; j >= 0; j--) 
        { 
            if (mat[N-1][j] > maxv) 
                maxv = mat[N - 1][j]; 
            maxArr[N-1][j] = maxv; 
        } 
       
        // preprocess last column 
        maxv = mat[N - 1][N - 1];  // Initialize max 
        for (int i = N - 2; i >= 0; i--) 
        { 
            if (mat[i][N - 1] > maxv) 
                maxv = mat[i][N - 1]; 
            maxArr[i][N - 1] = maxv; 
        } 
       
        // preprocess rest of the matrix from bottom 
        for (int i = N-2; i >= 0; i--) 
        { 
            for (int j = N-2; j >= 0; j--) 
            { 
                // Update maxValue 
                if (maxArr[i+1][j+1] - mat[i][j] > maxValue) 
                    maxValue = maxArr[i + 1][j + 1] - mat[i][j]; 
       
                // set maxArr (i, j) 
                maxArr[i][j] = Math.max(mat[i][j], 
                                   Math.max(maxArr[i][j + 1], 
                                       maxArr[i + 1][j]) ); 
            } 
        } 
       
        return maxValue; 
    } 
      
    // Driver code 
    public static void main (String[] args)  
    { 
        int N = 5; 
  
        int mat[][] = { 
                      { 1, 2, -1, -4, -20 }, 
                      { -8, -3, 4, 2, 1 }, 
                      { 3, 8, 6, 1, 3 }, 
                      { -4, -1, 1, 7, -6 }, 
                      { 0, -4, 10, -5, 1 } 
                   }; 
  
        System.out.print("Maximum Value is " +  
                           findMaxValue(N,mat)); 
    } 
} 
   
// Contributed by Prakriti Gupta 

Python3

# An efficient method to find maximum value  
# of mat[d] - ma[a][b] such that c > a and d > b 
  
import sys 
N = 5
  
# The function returns maximum value  
# A(c,d) - A(a,b) over all choices of  
# indexes such that both c > a and d > b. 
def findMaxValue(mat): 
  
    # stores maximum value 
    maxValue = -sys.maxsize -1
  
    # maxArr[i][j] stores max of elements  
    # in matrix from (i, j) to (N-1, N-1) 
    maxArr = [[0 for x in range(N)] 
                 for y in range(N)] 
  
    # last element of maxArr will be  
    # same's as of the input matrix 
    maxArr[N - 1][N - 1] = mat[N - 1][N - 1] 
  
    # preprocess last row 
    maxv = mat[N - 1][N - 1]; # Initialize max 
    for j in range (N - 2, -1, -1): 
      
        if (mat[N - 1][j] > maxv): 
            maxv = mat[N - 1][j] 
        maxArr[N - 1][j] = maxv 
      
    # preprocess last column 
    maxv = mat[N - 1][N - 1] # Initialize max 
    for i in range (N - 2, -1, -1): 
      
        if (mat[i][N - 1] > maxv): 
            maxv = mat[i][N - 1] 
        maxArr[i][N - 1] = maxv 
  
    # preprocess rest of the matrix 
    # from bottom 
    for i in range (N - 2, -1, -1): 
      
        for j in range (N - 2, -1, -1): 
          
            # Update maxValue 
            if (maxArr[i + 1][j + 1] -
                mat[i][j] > maxValue): 
                maxValue = (maxArr[i + 1][j + 1] - 
                                       mat[i][j]) 
  
            # set maxArr (i, j) 
            maxArr[i][j] = max(mat[i][j],  
                           max(maxArr[i][j + 1],  
                               maxArr[i + 1][j])) 
          
    return maxValue 
  
# Driver Code 
mat = [[ 1, 2, -1, -4, -20 ], 
       [-8, -3, 4, 2, 1 ], 
       [ 3, 8, 6, 1, 3 ], 
       [ -4, -1, 1, 7, -6] , 
       [0, -4, 10, -5, 1 ]] 
                      
print ("Maximum Value is", 
        findMaxValue(mat)) 
  
# This code is contributed by iAyushRaj 

C#

// An efficient method to find  
// maximum value of mat1[d] 
// - ma[a][b] such that c > a  
// and d > b 
using System; 
class GFG  { 
      
    // The function returns 
    // maximum value A(c,d) - A(a,b) 
    // over all choices of indexes  
    // such that both c > a 
    // and d > b. 
    static int findMaxValue(int N, int [,]mat) 
    { 
          
        //stores maximum value 
        int maxValue = int.MinValue; 
      
        // maxArr[i][j] stores max  
        // of elements in matrix 
        // from (i, j) to (N-1, N-1) 
        int [,]maxArr = new int[N, N]; 
      
        // last element of maxArr  
        // will be same's as of 
        // the input matrix 
        maxArr[N - 1, N - 1] = mat[N - 1,N - 1]; 
      
        // preprocess last row 
         // Initialize max 
        int maxv = mat[N - 1, N - 1]; 
        for (int j = N - 2; j >= 0; j--) 
        { 
            if (mat[N - 1, j] > maxv) 
                maxv = mat[N - 1, j]; 
            maxArr[N - 1, j] = maxv; 
        } 
      
        // preprocess last column 
        // Initialize max 
        maxv = mat[N - 1,N - 1];  
        for (int i = N - 2; i >= 0; i--) 
        { 
            if (mat[i, N - 1] > maxv) 
                maxv = mat[i,N - 1]; 
            maxArr[i,N - 1] = maxv; 
        } 
      
        // preprocess rest of the 
        // matrix from bottom 
        for (int i = N - 2; i >= 0; i--) 
        { 
            for (int j = N - 2; j >= 0; j--) 
            { 
                  
                // Update maxValue 
                if (maxArr[i + 1,j + 1] -  
                     mat[i, j] > maxValue) 
                    maxValue = maxArr[i + 1,j + 1] -  
                                         mat[i, j]; 
      
                // set maxArr (i, j) 
                maxArr[i,j] = Math.Max(mat[i, j], 
                              Math.Max(maxArr[i, j + 1], 
                              maxArr[i + 1, j]) ); 
            } 
        } 
      
        return maxValue; 
    } 
      
    // Driver code 
    public static void Main ()  
    { 
        int N = 5; 
  
        int [,]mat = {{ 1, 2, -1, -4, -20 }, 
                      { -8, -3, 4, 2, 1 }, 
                      { 3, 8, 6, 1, 3 }, 
                      { -4, -1, 1, 7, -6 }, 
                      { 0, -4, 10, -5, 1 }}; 
        Console.Write("Maximum Value is " +  
                        findMaxValue(N,mat)); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php  
// An efficient method to find  
// maximum value of mat[d] - ma[a][b]  
// such that c > a and d > b 
$N = 5; 
  
// The function returns maximum  
// value A(c,d) - A(a,b) over 
// all choices of indexes such  
// that both c > a and d > b. 
function findMaxValue($mat) 
{ 
    global $N; 
      
    // stores maximum value 
    $maxValue = PHP_INT_MIN; 
  
    // maxArr[i][j] stores max  
    // of elements in matrix 
    // from (i, j) to (N-1, N-1) 
    $maxArr[$N][$N] = array(); 
  
    // last element of maxArr  
    // will be same's as of 
    // the input matrix 
    $maxArr[$N - 1][$N - 1] = $mat[$N - 1][$N - 1]; 
  
    // preprocess last row 
    $maxv = $mat[$N - 1][$N - 1]; // Initialize max 
    for ($j = $N - 2; $j >= 0; $j--) 
    { 
        if ($mat[$N - 1][$j] > $maxv) 
            $maxv = $mat[$N - 1][$j]; 
        $maxArr[$N - 1][$j] = $maxv; 
    } 
  
    // preprocess last column 
    $maxv = $mat[$N - 1][$N - 1]; // Initialize max 
    for ($i = $N - 2; $i >= 0; $i--) 
    { 
        if ($mat[$i][$N - 1] > $maxv) 
            $maxv = $mat[$i][$N - 1]; 
        $maxArr[$i][$N - 1] = $maxv; 
    } 
  
    // preprocess rest of the 
    // matrix from bottom 
    for ($i = $N - 2; $i >= 0; $i--) 
    { 
        for ($j = $N - 2; $j >= 0; $j--) 
        { 
            // Update maxValue 
            if ($maxArr[$i + 1][$j + 1] -  
                $mat[$i][$j] > $maxValue) 
                $maxValue = $maxArr[$i + 1][$j + 1] -  
                            $mat[$i][$j]; 
  
            // set maxArr (i, j) 
            $maxArr[$i][$j] = max($mat[$i][$j], 
                              max($maxArr[$i][$j + 1], 
                                  $maxArr[$i + 1][$j])); 
        } 
    } 
  
    return $maxValue; 
} 
  
// Driver Code 
$mat = array(array(1, 2, -1, -4, -20), 
             array(-8, -3, 4, 2, 1), 
             array(3, 8, 6, 1, 3), 
             array(-4, -1, 1, 7, -6), 
             array(0, -4, 10, -5, 1) 
                    ); 
echo "Maximum Value is ".  
      findMaxValue($mat); 
  
// This code is contributed  
// by ChitraNayal 
?> 

Output:

Maximum Value is 18

If we are allowed to modify of the matrix, we can avoid using extra space and use input matrix instead.

Exercise: Print index (a, b) and (c, d) as well.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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