# 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

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.

 // 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[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 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 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# 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)

 = 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

The 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.

 // 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][0],                    &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[0]);   cout << maximumSum(arr, n);   return 0; }

 // 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   // previoulsy 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 Chitranayal

 # 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)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :