Maximum sum of increasing order elements from n arrays
Last Updated :
12 Oct, 2022
Given n arrays of size m each. Find the maximum sum obtained by selecting a number from each array such that the elements selected from the i-th array are more than the element selected from (i-1)-th array. If maximum sum cannot be obtained then return 0.
Examples:
Input : arr[][] = {{1, 7, 3, 4},
{4, 2, 5, 1},
{9, 5, 1, 8}}
Output : 18
Explanation :
We can select 4 from first array, 5 from
second array and 9 from third array.
Input : arr[][] = {{9, 8, 7},
{6, 5, 4},
{3, 2, 1}}
Output : 0
The idea is to start picking from the last array. We pick the maximum element from the last array, then we move to the second last array. In the second last array, we find the largest element which is smaller than the maximum element picked from the last array. We repeat this process until we reach the first array.
To obtain maximum sum we can sort all arrays and start bottom to up traversing each array from right to left and choose a number such that it is greater than the previous element. If we are not able to select an element from the array then return 0.
C++
// CPP program to find maximum sum
// by selecting a element from n arrays
#include <bits/stdc++.h>
#define M 4
using namespace std;
// To calculate maximum sum by
// selecting element from each array
int maximumSum(int a[][M], int n) {
// Sort each array
for (int i = 0; i < n; i++)
sort(a[i], a[i] + M);
// Store maximum element
// of last array
int sum = a[n - 1][M - 1];
int prev = a[n - 1][M - 1];
int i, j;
// Selecting maximum element from
// previously selected element
for (i = n - 2; i >= 0; i--) {
for (j = M - 1; j >= 0; j--) {
if (a[i][j] < prev) {
prev = a[i][j];
sum += prev;
break;
}
}
// j = -1 means no element is
// found in a[i] so return 0
if (j == -1)
return 0;
}
return sum;
}
// Driver program to test maximumSum
int main() {
int arr[][M] = {{1, 7, 3, 4},
{4, 2, 5, 1},
{9, 5, 1, 8}};
int n = sizeof(arr) / sizeof(arr[0]);
cout << maximumSum(arr, n);
return 0;
}
// Java program to find
// maximum sum by selecting
// a element from n arrays
import java.io.*;
class GFG
{
static int M = 4;
static int arr[][] = {{1, 7, 3, 4},
{4, 2, 5, 1},
{9, 5, 1, 8}};
static void sort(int a[][],
int row, int n)
{
for (int i = 0; i < M - 1; i++)
{
if(a[row][i] > a[row][i + 1])
{
int temp = a[row][i];
a[row][i] = a[row][i + 1];
a[row][i + 1] = temp;
}
}
}
// To calculate maximum
// sum by selecting element
// from each array
static int maximumSum(int a[][],
int n)
{
// Sort each array
for (int i = 0; i < n; i++)
sort(a, i, n);
// Store maximum element
// of last array
int sum = a[n - 1][M - 1];
int prev = a[n - 1][M - 1];
int i, j;
// Selecting maximum element
// from previously selected
// element
for (i = n - 2; i >= 0; i--)
{
for (j = M - 1; j >= 0; j--)
{
if (a[i][j] < prev)
{
prev = a[i][j];
sum += prev;
break;
}
}
// j = -1 means no element
// is found in a[i] so
// return 0
if (j == -1)
return 0;
}
return sum;
}
// Driver Code
public static void main(String args[])
{
int n = arr.length;
System.out.print(maximumSum(arr, n));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
# Python3 program to find
# maximum sum by selecting
# a element from n arrays
M = 4;
# To calculate maximum sum
# by selecting element from
# each array
def maximumSum(a, n) :
global M;
# Sort each array
for i in range(0, n) :
a[i].sort();
# Store maximum element
# of last array
sum = a[n - 1][M - 1];
prev = a[n - 1][M - 1];
# Selecting maximum
# element from previously
# selected element
for i in range(n - 2,
-1, -1) :
for j in range(M - 1,
-1, -1) :
if (a[i][j] < prev) :
prev = a[i][j];
sum += prev;
break;
# j = -1 means no element
# is found in a[i] so
# return 0
if (j == -1) :
return 0;
return sum;
# Driver Code
arr = [[1, 7, 3, 4],
[4, 2, 5, 1],
[9, 5, 1, 8]];
n = len(arr) ;
print (maximumSum(arr, n));
# This code is contributed by
# Manish Shaw(manishshaw1)
// C# program to find maximum
// sum by selecting a element
// from n arrays
using System;
class GFG
{
static int M = 4;
static void sort(ref int[,] a,
int row, int n)
{
for (int i = 0; i < M-1; i++)
{
if(a[row, i] > a[row, i + 1])
{
int temp = a[row, i];
a[row, i] = a[row, i + 1];
a[row, i + 1] = temp;
}
}
}
// To calculate maximum
// sum by selecting
// element from each array
static int maximumSum(int[,] a,
int n)
{
int i = 0, j = 0;
// Sort each array
for (i = 0; i < n; i++)
sort(ref a, i, n);
// Store maximum element
// of last array
int sum = a[n - 1, M - 1];
int prev = a[n - 1, M - 1];
// Selecting maximum element
// from previously selected
// element
for (i = n - 2; i >= 0; i--)
{
for (j = M - 1; j >= 0; j--)
{
if (a[i, j] < prev)
{
prev = a[i, j];
sum += prev;
break;
}
}
// j = -1 means no element
// is found in a[i] so
// return 0
if (j == -1)
return 0;
}
return sum;
}
// Driver Code
static void Main()
{
int [,]arr = new int[,]{{1, 7, 3, 4},
{4, 2, 5, 1},
{9, 5, 1, 8}};
int n = arr.GetLength(0);
Console.Write(maximumSum(arr, n));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)
<?php
// PHP program to find maximum
// sum by selecting a element
// from n arrays
$M = 4;
// To calculate maximum sum
// by selecting element from
// each array
function maximumSum($a, $n)
{
global $M;
// Sort each array
for ($i = 0; $i < $n; $i++)
sort($a[$i]);
// Store maximum element
// of last array
$sum = $a[$n - 1][$M - 1];
$prev = $a[$n - 1][$M - 1];
$i; $j;
// Selecting maximum element from
// previously selected element
for ($i = $n - 2; $i >= 0; $i--)
{
for ($j = $M - 1; $j >= 0; $j--)
{
if ($a[$i][$j] < $prev)
{
$prev = $a[$i][$j];
$sum += $prev;
break;
}
}
// j = -1 means no element is
// found in a[i] so return 0
if ($j == -1)
return 0;
}
return $sum;
}
// Driver Code
$arr = array(array(1, 7, 3, 4),
array(4, 2, 5, 1),
array(9, 5, 1, 8));
$n = sizeof($arr) ;
echo maximumSum($arr, $n);
// This code is contributed by m_kit
?>
<script>
// JavaScript program to find
// maximum sum by selecting
// a element from n arrays
let M = 4;
// To calculate maximum sum
// by selecting element from
// each array
function maximumSum(a, n)
{
// Store maximum element
// of last array
let prev = Math.max(a[n - 1][0],
a[n - 1][M - 1] + 1);
// Selecting maximum element from
// previously selected element
let sum = prev;
for (let i = n - 2; i >= 0; i--)
{
let max_smaller = Number.MIN_VALUE;
for (let j = M - 1; j >= 0; j--)
{
if (a[i][j] < prev &&
a[i][j] > max_smaller)
max_smaller = a[i][j];
}
// max_smaller equals to
// INT_MIN means no element
// is found in a[i] so return 0
if (max_smaller == Number.MIN_VALUE)
return 0;
prev = max_smaller;
sum += max_smaller;
}
return sum;
}
// Driver code
let arr = [[1, 7, 3, 4],
[4, 2, 5, 1],
[9, 5, 1, 8]];
let n = arr.length;
document.write(maximumSum(arr, n));
</script>
The Worst-Case Time Complexity: O(mn Log m)
Auxiliary Space: O(1)
We can optimize the above solution to work in O(mn). We can skip sorting to find the maximum elements.
C++
// CPP program to find maximum sum by selecting a element
// from n arrays
#include <bits/stdc++.h>
#define M 4
using namespace std;
// To calculate maximum sum by selecting element from each
// array
int maximumSum(int a[][M], int n)
{
// Store maximum element of last array
int prev
= *max_element(&a[n - 1][0], &a[n - 1][M - 1] + 1);
// Selecting maximum element from previously selected
// element
int sum = prev;
for (int i = n - 2; i >= 0; i--) {
int max_smaller = INT_MIN;
for (int j = M - 1; j >= 0; j--)
if (a[i][j] < prev && a[i][j] > max_smaller)
max_smaller = a[i][j];
// max_smaller equals to INT_MIN means no element is
// found in a[i] so return 0
if (max_smaller == INT_MIN)
return 0;
prev = max_smaller;
sum += max_smaller;
}
return sum;
}
// Driver program to test maximumSum
int main()
{
int arr[][M] = { { 1, 7, 3, 4 },
{ 4, 2, 5, 1 },
{ 9, 5, 1, 8 } };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maximumSum(arr, n);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// C program to find maximum sum by selecting a element from
// n arrays
#include <limits.h>
#include <stdio.h>
#define M 4
// To calculate maximum sum by selecting element from each
// array
int maximumSum(int a[][M], int n)
{
// Store maximum element of last array
int prev = INT_MAX;
// Selecting maximum element from previously selected
// element
int sum = 0;
for (int i = n - 1; i >= 0; i--) {
int max_smaller = INT_MIN;
for (int j = 0; j < M; j++)
if (a[i][j] > max_smaller && a[i][j] < prev)
max_smaller = a[i][j];
// max_smaller equals to INT_MIN means no element is
// found in a[i] so return 0
if (max_smaller == INT_MIN)
return 0;
prev = max_smaller;
sum += max_smaller;
}
return sum;
}
// Driver program to test maximumSum
int main()
{
int arr[][M] = { { 1, 7, 3, 4 },
{ 4, 2, 5, 1 },
{ 9, 5, 1, 8 } };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", maximumSum(arr, n));
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// Java program to find maximum sum by selecting a element
// from n arrays
import java.util.*;
class GFG {
static int M = 4;
// To calculate maximum sum by selecting element from
// each array
static int maximumSum(int a[][], int n)
{
// Store maximum element of last array
int prev
= Math.max(a[n - 1][0], a[n - 1][M - 1] + 1);
// Selecting maximum element from previously
// selected element
int sum = prev;
for (int i = n - 2; i >= 0; i--) {
int max_smaller = Integer.MIN_VALUE;
for (int j = M - 1; j >= 0; j--)
if (a[i][j] < prev && a[i][j] > max_smaller)
max_smaller = a[i][j];
// max_smaller equals to INT_MIN means no
// element is found in a[i] so return 0
if (max_smaller == Integer.MIN_VALUE)
return 0;
prev = max_smaller;
sum += max_smaller;
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[][] = { { 1, 7, 3, 4 },
{ 4, 2, 5, 1 },
{ 9, 5, 1, 8 } };
int n = arr.length;
System.out.print(maximumSum(arr, n));
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# Python3 program to find maximum sum
# by selecting a element from n arrays
M = 4
# To calculate maximum sum by
# selecting element from each array
def maximumSum(a, n):
# Store maximum element of last array
prev = max(max(a))
# Selecting maximum element from
# previously selected element
Sum = prev
for i in range(n - 2, -1, -1):
max_smaller = -10**9
for j in range(M - 1, -1, -1):
if (a[i][j] < prev and
a[i][j] > max_smaller):
max_smaller = a[i][j]
# max_smaller equals to INT_MIN means
# no element is found in a[i] so
# return 0
if (max_smaller == -10**9):
return 0
prev = max_smaller
Sum += max_smaller
return Sum
# Driver Code
arr = [[1, 7, 3, 4],
[4, 2, 5, 1],
[9, 5, 1, 8]]
n = len(arr)
print(maximumSum(arr, n))
# This code is contributed by mohit kumar
// C# program to find
// maximum sum by selecting
// a element from n arrays
using System;
class GFG
{
static int M = 4;
// To calculate maximum sum
// by selecting element from
// each array
static int maximumSum(int[,] a, int n)
{
// Store maximum element
// of last array
int prev = Math.Max(a[n - 1, 0],
a[n - 1, M - 1] + 1);
// Selecting maximum element from
// previously selected element
int sum = prev;
for(int i = n - 2; i >= 0; i--)
{
int max_smaller = Int32.MinValue;
for(int j = M - 1; j >= 0; j--)
{
if(a[i, j] < prev && a[i, j] > max_smaller)
{
max_smaller = a[i, j];
}
}
// max_smaller equals to
// INT_MIN means no element
// is found in a[i] so return 0
if(max_smaller == Int32.MinValue)
{
return 0;
}
prev = max_smaller;
sum += max_smaller;
}
return sum;
}
// Driver code
static public void Main ()
{
int[,] arr = {{1, 7, 3, 4},{4, 2, 5, 1},{9, 5, 1, 8}};
int n = arr.GetLength(0);
Console.Write(maximumSum(arr, n));
}
}
// This code is contributed by avanitrachhadiya2155
<script>
// JavaScript program to find
// maximum sum by selecting
// a element from n arrays
let M = 4;
// To calculate maximum sum
// by selecting element from
// each array
function maximumSum(a, n)
{
// Store maximum element
// of last array
let prev = Math.max(a[n - 1][0],
a[n - 1][M - 1] + 1);
// Selecting maximum element from
// previously selected element
let sum = prev;
for (let i = n - 2; i >= 0; i--)
{
let max_smaller = Number.MIN_VALUE;
for (let j = M - 1; j >= 0; j--)
{
if (a[i][j] < prev &&
a[i][j] > max_smaller)
max_smaller = a[i][j];
}
// max_smaller equals to
// INT_MIN means no element
// is found in a[i] so return 0
if (max_smaller == Number.MIN_VALUE)
return 0;
prev = max_smaller;
sum += max_smaller;
}
return sum;
}
// Driver code
let arr = [[1, 7, 3, 4],
[4, 2, 5, 1],
[9, 5, 1, 8]];
let n = arr.length;
document.write(maximumSum(arr, n));
// This code is contributed by souravghosh0416.
</script>
Time Complexity: O(mn)
Auxiliary Space: O(1)
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