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 functionint 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 > bimport 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 > bN = 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 Codemat = [[ 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 > busing 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?> |
Javascript
<script>// A Naive method to find maximum value of mat1[d][e]// - ma[a][b] such that d > a and e > b // 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(N,mat) { // stores maximum value let maxValue = Number.MIN_VALUE; // Consider all possible pairs mat[a][b] and // mat1[d][e] for (let a = 0; a < N - 1; a++) for (let b = 0; b < N - 1; b++) for (let d = a + 1; d < N; d++) for (let 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 let N = 5; let 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 ]]; document.write("Maximum Value is " +findMaxValue(N,mat)); // This code is contributed by rag2127</script> |
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 functionint 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 > bimport 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 > bimport sysN = 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 Codemat = [[ 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 > busing 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?> |
Javascript
<script>// An efficient method to find maximum value of mat1[d]// - ma[a][b] such that c > a and d > b // 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(N,mat) { // stores maximum value let maxValue = Number.MIN_VALUE; // maxArr[i][j] stores max of elements in matrix // from (i, j) to (N-1, N-1) let maxArr=new Array(N); for(let i = 0; i < N; i++) { maxArr[i]=new Array(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 let maxv = mat[N-1][N-1]; // Initialize max for (let 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 (let 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 (let i = N-2; i >= 0; i--) { for (let 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 let N = 5; let 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 ]]; document.write("Maximum Value is " + findMaxValue(N,mat)); // This code is contributed by avanitrachhadiya2155</script> |
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.
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