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
Time complexity: O(N4).
Auxiliary Space: O(1)
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.
Implementation:
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
Time complexity: O(N2).
Auxiliary Space: O(N2)
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.
An optimal approach is with space complexity O(N).
Instead of using the maxArr matrix, we can use two separate vectors (temp1 and temp2) to get maxArr[i+1][j] and maxArr[i][j+1] values.
C++
// An optimal 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]){ vector<int> temp1(N), temp2(N); temp1[N - 1] = mat[N - 1][N - 1]; // Fill temp1 for (int j = N - 2; j >= 0; j--) temp1[j] = max(temp1[j + 1], mat[N - 1][j]); // stores maximum value int maxValue = INT_MIN; // Iterate over the remaining rows for (int i = N - 2; i >= 0; i--) { // Initialize the last element of temp2 temp2[N - 1] = max(temp1[N - 1], mat[i][N - 1]); for (int j = N - 2; j >= 0; j--) { // update temp2 and maxValue maxValue = max(maxValue, temp1[j + 1] - mat[i][j]); temp2[j] = max( { mat[i][j], temp1[j], temp2[j + 1] }); } // Set temp1 to temp2 for the next iteration temp1 = temp2; } // Return the maximum value 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;}// This code is contributed by Tapesh(tapeshdua420) |
Output
Maximum Value is 18
The time complexity of the findMaxValue() function is O(N^2) because it iterates over each element of the matrix exactly once.
The space complexity of the function is O(N) because it uses two vectors of size N (temp1 and temp2) to store the maximum values seen so far in each row and column.
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