# Maximum sum of increasing order elements from n arrays

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

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to start picking from last array. We pick the maximum element from last array, then we move to second last array. In second last array, we find the largest element which is smaller than the maximum element picked from last array. We repeat this process till we reach 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 array then return 0.

## C++

 `// CPP program to find maximum sum ` `// by selecting a element from n arrays ` `#include ` `#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 ` `  ``// previoulsy 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); ` `  ``cout << maximumSum(arr, n); ` `  ``return` `0; ` `} `

## Java

 `// 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 previoulsy 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

 `# 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 previoulsy ` `    ``# 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#

 `// 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 previoulsy 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

 `= 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 ` `?> `

Output:

```18
```

Worst Case Time Complexity : O(mn Log m)

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 ` `#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], ` `                   ``&a[n-1][M-1] + 1); ` ` `  `  ``// Selecting maximum element from ` `  ``// previoulsy 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); ` `  ``cout << maximumSum(arr, n); ` `  ``return` `0; ` `} `

## Python3

 `# 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 ` `    ``# previoulsy 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 `

Output:

```18
```

Time Complexity: O(mn)

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