Minimum sum of multiplications of n numbers

Given n integers. The task is to minimize the sum of multiplication of all the numbers by taking two adjacent numbers at a time and putting back their sum % 100 till a number is left.

Examples :

Input : 40 60 20 
Output : 2400  
Explanation: There are two possible cases:
1st possibility: Take 40 and 60, so multiplication=2400
and put back (60+40) % 100 = 0, making it 0, 20.
Multiplying 0 and 20 we get 0 so 
multiplication = 2400+0 = 2400. Put back (0+20)%100 = 20. 
2nd possibility: take 60 and 20, so 60*20 = 1200,
put back (60+20)%100 = 80, making it [40, 80] 
multiply 40*80 to get 3200, so multiplication
sum = 1200+3200 = 4400. Put back (40+80)%100 = 20 

Input : 5 6 
Output : 30 
Explanation: Only possibility is 5*6=30

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

Approach: The problem is a variation of Matrix chain Multiplication Dynamic Programming

The idea is to partition N numbers into every possible value of k. Solve recursively for smaller parts and add the multiplication and store the minimum of them. Since we are dividing it into k parts, for every DP_i,j we will have k partitions i<=k<j , store the minimum of them. So we get the formula similar to Matrix chain Multiplication Dynamic Programming.

DP_i,j = min(DP_i,j , (DP_i,k+DP_k+1,j+(cumulative_sum_i,k*cumulative_sum_k+1,j) ) )
for every i<=k<j.

Since many subproblems will be repeating, hence we use memoization to store the values in a nXn matrix.

Given below is the illustration of the above approach:

C++

// CPP program to find the  
// minimum sum of multiplication 
// of n numbers 
#include <bits/stdc++.h> 
using namespace std; 
  
// Used in recursive  
// memoized solution 
long long dp[1000][1000]; 
  
// function to calculate the cumulative  
// sum from a[i] to a[j] 
long long sum(int a[], int i, int j) 
{      
    long long ans = 0; 
    for (int m = i; m <= j; m++)      
        ans = (ans + a[m]) % 100; 
    return ans; 
} 
  
long long solve(int a[], int i, int j) 
{ 
    // base case  
    if (i == j) 
        return 0;  
      
    // memoization, if the partition  
    // has been called before then  
    // return the stored value  
    if (dp[i][j] != -1) 
        return dp[i][j]; 
      
    // store a max value  
    dp[i][j] = INT_MAX; 
      
    // we break them into k partitions  
    for (int k = i; k < j; k++) 
    { 
        // store the min of the  
        // formula thus obtained 
        dp[i][j] = min(dp[i][j], (solve(a, i, k) + 
                              solve(a, k + 1, j) +  
              (sum(a, i, k) * sum(a, k + 1, j)))); 
    } 
      
    // return the minimum  
    return dp[i][j]; 
} 
  
void intialize(int n) 
{ 
    for (int i = 0; i <= n; i++)  
        for (int j = 0; j <= n; j++) 
            dp[i][j] = -1;      
} 
  
// Driver code  
int main() { 
    int a[] = {40, 60, 20};  
    int n = sizeof(a) / sizeof(a[0]); 
    intialize(n); 
    cout << solve(a, 0, n - 1) << endl; 
    return 0; 
} 

Java

// Java program to find the   
// minimum sum of multiplication 
// of n numbers 
import java.io.*; 
import java.math.*; 
  
  
class GFG 
{ 
      
    // Used in recursive 
    // memoized solution 
    static long dp[][] = new long[1000][1000]; 
      
    // function to calculate the  
    // cumulative  sum from a[i] to a[j] 
    static long sum(int a[], int i, int j) 
    {      
        long ans = 0; 
        for (int m = i; m <= j; m++)      
            ans = (ans + a[m]) % 100; 
        return ans; 
    } 
      
    static long solve(int a[], int i, int j) 
    { 
        // base case  
        if (i == j) 
            return 0;  
          
        // memoization, if the partition  
        // has been called before then  
        // return the stored value  
        if (dp[i][j] != -1) 
            return dp[i][j]; 
          
        // store a max value  
        dp[i][j] = 100000000; 
          
        // we break them into k partitions  
        for (int k = i; k < j; k++) 
        { 
            // store the min of the 
            // formula thus obtained 
            dp[i][j] = Math.min(dp[i][j], (solve(a, i, k) + 
                                       solve(a, k + 1, j) +  
                        (sum(a, i, k) * sum(a, k + 1,j)))); 
        } 
          
        // return the minimum  
        return dp[i][j]; 
    } 
      
    static void intialize(int n) 
    { 
        for (int i = 0; i <= n; i++)  
            for (int j = 0; j <= n; j++) 
                dp[i][j] = -1;      
    } 
      
    // Driver code  
    public static void main(String args[])  
    { 
        int a[] = {40, 60, 20};  
        int n = a.length; 
        intialize(n); 
        System.out.println(solve(a, 0, n - 1)); 
          
    } 
} 
  
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 program to find the  
# minimum sum of multiplication  
# of n numbers  
  
import numpy as np 
import sys 
  
# Used in recursive  
# memoized solution  
dp = np.zeros((1000,1000))  
  
# function to calculate the cumulative  
# sum from a[i] to a[j]  
def sum(a, i, j) : 
          
    ans = 0
    for m in range(i, j + 1) :  
        ans = (ans + a[m]) % 100
          
    return ans 
  
  
def solve(a, i, j) : 
  
    # base case  
    if (i == j) :  
        return 0
      
    # memoization, if the partition  
    # has been called before then  
    # return the stored value  
    if (dp[i][j] != -1) : 
                  
        return dp[i][j]  
      
    # store a max value  
    dp[i][j] = sys.maxsize 
      
    # we break them into k partitions  
    for k in range(i, j) : 
                  
        # store the min of the  
        # formula thus obtained  
        dp[i][j] = min(dp[i][j], (solve(a, i, k) + solve(a, k + 1, j)  
                                + (sum(a, i, k) * sum(a, k + 1, j))))  
      
    # return the minimum  
    return dp[i][j] 
  
  
def intialize(n) : 
          
    for i in range(n + 1) : 
        for j in range(n + 1) : 
            dp[i][j] = -1    
  
#Driver code  
if __name__ == "__main__" : 
          
    a = [40, 60, 20] 
    n = len(a)  
    intialize(n) 
    print(int(solve(a, 0, n - 1)))  
  
# This code is contributed by Ryuga 

C#

// C# program to find the  
// minimum sum of multiplication  
// of n numbers 
using System; 
  
class GFG  
{ 
      
    // Used in recursive  
    // memoized solution 
    static long [,]dp = new long[1000, 1000]; 
      
    // Function to calculate the cumulative  
    // sum from a[i] to a[j] 
    static long sum(int []a, int i, int j) 
    {  
        long ans = 0; 
        for (int m = i; m <= j; m++)  
            ans = (ans + a[m]) % 100; 
        return ans; 
    } 
      
    static long solve(int []a, int i, int j) 
    { 
        // base case  
        if (i == j) 
            return 0;  
          
        // memoization, if the partition  
        // has been called before then  
        // return the stored value  
        if (dp[i, j] != -1) 
            return dp[i, j]; 
          
        // store a max value  
        dp[i, j] = 100000000; 
          
        // we break them into k partitions  
        for (int k = i; k < j; k++) 
        { 
            // store the min of the  
            // formula thus obtained 
            dp[i, j] = Math.Min(dp[i, j], (solve(a, i, k) + 
                                       solve(a, k + 1, j) +  
                       (sum(a, i, k) * sum(a, k + 1, j)))); 
        } 
          
        // return the minimum  
        return dp[i, j]; 
    } 
      
    static void intialize(int n) 
    { 
        for (int i = 0; i <= n; i++)  
            for (int j = 0; j <= n; j++) 
                dp[i, j] = -1;  
    } 
      
    // Driver code  
    public static void Main()  
    { 
        int []a = {40, 60, 20};  
        int n = a.Length; 
        intialize(n); 
        Console.WriteLine(solve(a, 0, n - 1)); 
          
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find the  
// minimum sum of multiplication  
// of n numbers 
  
  
// Used in recursive 
// memoized solution 
$dp = array(array()); 
  
// function to calculate the  
// cumulative sum from a[i] to a[j] 
function sum( $a, $i, $j) 
{  
    $ans = 0; 
    for ( $m = $i; $m <= $j; $m++)  
        $ans = ($ans + $a[$m]) % 100; 
    return $ans; 
} 
  
function solve( $a, $i, $j) 
{ 
    global $dp; 
    // base case  
    if ($i == $j) 
        return 0;  
      
    // memoization, if the partition 
    // has been  called before then 
    // return the stored value  
    if ($dp[$i][$j] != -1) 
        return $dp[$i][$j]; 
      
    // store a max value  
    $dp[$i][$j] = PHP_INT_MAX; 
      
    // we break them into 
    // k partitions  
    for ( $k = $i; $k < $j; $k++) 
    { 
        // store the min of the 
        // formula thus obtained 
        $dp[$i][$j] = min($dp[$i][$j],  
                      (solve($a, $i, $k) + 
                       solve($a, $k + 1, $j) +  
                      (sum($a, $i, $k) *  
                       sum($a, $k + 1, $j)))); 
    } 
      
    // return the minimum  
    return $dp[$i][$j]; 
} 
  
function intialize( $n) 
{ 
    global $dp; 
    for ( $i = 0; $i <= $n; $i++)  
        for ( $j = 0; $j <= $n; $j++) 
            $dp[$i][$j] = -1;  
} 
  
// Driver code  
$a = array(40, 60, 20);  
$n = count($a); 
intialize($n); 
echo solve($a, 0, $n - 1) ; 
  
// This code is contributed by anuj_67. 
?> 

Output :

Time Complexity: O(n^3)
Auxiliary Space: O(n^2)

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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