Given an array a[] consisting of N integers and an integer K, the task is to find the minimum cost to reduce the given array to a single element by choosing any pair of consecutive array elements and replace them by (a[i] + a[i+1]) for a cost K * (a[i] + a[i+1]).
Examples:
Input: a[] = {1, 2, 3}, K = 2
Output: 18
Explanation:
Replacing {1, 2} by 3 modifies the array to {3, 3}. Cost 2 * 3 = 6
Replacing {3, 3} by 6 modifies the array to {6}. Cost 2 * 6 = 12
Therefore, the total cost is 18
Input: a[] = {4, 5, 6, 7}, K = 3
Output: 132
Naive Approach:
The simplest solution is to split the array into two halves, for every index and compute the cost of the two halves recursively and finally add their respective costs.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;#define inf 10000009// Function to combine the sum of the two halvesint Combine(int a[], int i, int j){ int sum = 0; // Calculate the sum from i to j for (int l = i; l <= j; l++) sum += a[l]; return sum;}// Function to minimize the cost to// reduce the array to a single elementint minCost(int a[], int i, int j, int k){ if (i >= j) { // Base case // If n = 1 or n = 0 return 0; } // Initialize cost to maximum value int best_cost = inf; // Iterate through all possible indices // and find the best index // to combine the subproblems for (int pos = i; pos < j; pos++) { // Compute left subproblem int left = minCost(a, i, pos, k); // Compute right subproblem int right = minCost(a, pos + 1, j, k); // Calculate the best cost best_cost = min(best_cost, left + right + k * Combine(a, i, j)); } // Return the answer return best_cost;}// Driver codeint main(){ int n = 4; int a[] = { 4, 5, 6, 7 }; int k = 3; cout << minCost(a, 0, n - 1, k) << endl; return 0;}// This code is contributed by PrinciRaj1992 |
Java
// Java Program to implement// the above approachimport java.io.*;class GFG { static int inf = 10000009; // Function to minimize the cost to // reduce the array to a single element public static int minCost(int a[], int i, int j, int k) { if (i >= j) { // Base case // If n = 1 or n = 0 return 0; } // Initialize cost to maximum value int best_cost = inf; // Iterate through all possible indices // and find the best index // to combine the subproblems for (int pos = i; pos < j; pos++) { // Compute left subproblem int left = minCost(a, i, pos, k); // Compute right subproblem int right = minCost(a, pos + 1, j, k); // Calculate the best cost best_cost = Math.min( best_cost, left + right + k * Combine(a, i, j)); } // Return the answer return best_cost; } // Function to combine the sum of the two halves public static int Combine(int a[], int i, int j) { int sum = 0; // Calculate the sum from i to j for (int l = i; l <= j; l++) sum += a[l]; return sum; } // Driver code public static void main(String[] args) { int n = 4; int a[] = { 4, 5, 6, 7 }; int k = 3; System.out.println(minCost(a, 0, n - 1, k)); }} |
Python3
# Python3 Program to implement# the above approachinf = 10000009;# Function to minimize the cost to# reduce the array to a single elementdef minCost(a, i, j, k): if (i >= j): # Base case # If n = 1 or n = 0 return 0; # Initialize cost to maximum value best_cost = inf; # Iterate through all possible indices # and find the best index # to combine the subproblems for pos in range(i, j): # Compute left subproblem left = minCost(a, i, pos, k); # Compute right subproblem right = minCost(a, pos + 1, j, k); # Calculate the best cost best_cost = min(best_cost, left + right + k * Combine(a, i, j)); # Return the answer return best_cost;# Function to combine # the sum of the two halvesdef Combine(a, i, j): sum = 0; # Calculate the sum from i to j for l in range(i, j + 1): sum += a[l]; return sum;# Driver codeif __name__ == '__main__': n = 4; a = [4, 5, 6, 7]; k = 3; print(minCost(a, 0, n - 1, k));# This code is contributed by Amit Katiyar |
C#
// C# Program to implement// the above approachusing System;class GFG{ static int inf = 10000009; // Function to minimize the cost to // reduce the array to a single element public static int minCost(int []a, int i, int j, int k) { if (i >= j) { // Base case // If n = 1 or n = 0 return 0; } // Initialize cost to maximum value int best_cost = inf; // Iterate through all possible indices // and find the best index // to combine the subproblems for (int pos = i; pos < j; pos++) { // Compute left subproblem int left = minCost(a, i, pos, k); // Compute right subproblem int right = minCost(a, pos + 1, j, k); // Calculate the best cost best_cost = Math.Min(best_cost, left + right + k * Combine(a, i, j)); } // Return the answer return best_cost; } // Function to combine the sum of the two halves public static int Combine(int []a, int i, int j) { int sum = 0; // Calculate the sum from i to j for (int l = i; l <= j; l++) sum += a[l]; return sum; } // Driver code public static void Main(String[] args) { int n = 4; int []a = { 4, 5, 6, 7 }; int k = 3; Console.WriteLine(minCost(a, 0, n - 1, k)); }}// This code is contributed by Rohit_ranjan |
Javascript
<script>// JavaScript Program to implement// the above approachvar inf = 10000009;// Function to combine the sum of the two halvesfunction Combine(a, i, j){ var sum = 0; // Calculate the sum from i to j for (var l = i; l <= j; l++) sum += a[l]; return sum;}// Function to minimize the cost to// reduce the array to a single elementfunction minCost(a, i, j, k){ if (i >= j) { // Base case // If n = 1 or n = 0 return 0; } // Initialize cost to maximum value var best_cost = inf; // Iterate through all possible indices // and find the best index // to combine the subproblems for (var pos = i; pos < j; pos++) { // Compute left subproblem var left = minCost(a, i, pos, k); // Compute right subproblem var right = minCost(a, pos + 1, j, k); // Calculate the best cost best_cost = Math.min(best_cost, left + right + k * Combine(a, i, j)); } // Return the answer return best_cost;}// Driver codevar n = 4;var a = [4, 5, 6, 7];var k = 3;document.write( minCost(a, 0, n - 1, k));</script> |
Output:
132
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use the concept of Dynamic Programming. Follow the steps below to solve the problem:
- Initialize a matrix dp[][] and such that dp[i][j] stores the sum from index i to j.
- Calculate sum(i, j) using Prefix Sum technique.
- Compute the sum of the two subproblems and update the cost with the minimum value.
- Store in dp[][] and return.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h>using namespace std;int inf = 10000000; // Function to generate the cost using // Prefix Sum Array technique vector<int> preprocess(vector<int> a, int n) { vector<int> p(n); p[0] = a[0]; for(int i = 1; i < n; i++) { p[i] = p[i - 1] + a[i]; } return p; }// Function to combine the sum of the// two subproblems int Combine(vector<int> p, int i, int j) { if (i == 0) return p[j]; else return p[j] - p[i - 1]; } // Function to minimize the cost to // add the array elements to a single element int minCost(vector<int> a, int i, int j, int k, vector<int> prefix, vector<vector<int>> dp) { if (i >= j) return 0; // Check if the value is // already stored in the array if (dp[i][j] != -1) return dp[i][j]; int best_cost = inf; for(int pos = i; pos < j; pos++) { // Compute left subproblem int left = minCost(a, i, pos, k, prefix, dp); // Compute left subproblem int right = minCost(a, pos + 1, j, k, prefix, dp); // Calculate minimum cost best_cost = min(best_cost, left + right + (k * Combine(prefix, i, j))); } // Store the answer to // avoid recalculation return dp[i][j] = best_cost; } // Driver code int main(){ int n = 4; vector<int> a = { 4, 5, 6, 7 }; int k = 3; // Initialise dp array vector<vector<int>> dp; dp.resize(n + 1, vector<int>(n + 1)); for(int i = 0; i < n + 1; i++) { for(int j = 0; j < n + 1; j++) { dp[i][j] = -1; } } // Preprocessing the array vector<int> prefix = preprocess(a, n); cout << minCost(a, 0, n - 1, k, prefix, dp) << endl; return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java Program for the above approachimport java.util.*;public class Main { static int inf = 10000000; // Function to minimize the cost to // add the array elements to a single element public static int minCost(int a[], int i, int j, int k, int[] prefix, int[][] dp) { if (i >= j) return 0; // Check if the value is // already stored in the array if (dp[i][j] != -1) return dp[i][j]; int best_cost = inf; for (int pos = i; pos < j; pos++) { // Compute left subproblem int left = minCost(a, i, pos, k, prefix, dp); // Compute left subproblem int right = minCost(a, pos + 1, j, k, prefix, dp); // Calculate minimum cost best_cost = Math.min( best_cost, left + right + (k * Combine(prefix, i, j))); } // Store the answer to // avoid recalculation return dp[i][j] = best_cost; } // Function to generate the cost using // Prefix Sum Array technique public static int[] preprocess(int[] a, int n) { int p[] = new int[n]; p[0] = a[0]; for (int i = 1; i < n; i++) p[i] = p[i - 1] + a[i]; return p; } // Function to combine the sum of the two subproblems public static int Combine(int[] p, int i, int j) { if (i == 0) return p[j]; else return p[j] - p[i - 1]; } // Driver Code public static void main(String args[]) { int n = 4; int a[] = { 4, 5, 6, 7 }; int k = 3; // Initialise dp array int dp[][] = new int[n + 1][n + 1]; for (int i[] : dp) Arrays.fill(i, -1); // Preprocessing the array int prefix[] = preprocess(a, n); System.out.println( minCost(a, 0, n - 1, k, prefix, dp)); }} |
Python3
# Python3 program for the above approachinf = 10000000# Function to minimize the cost to# add the array elements to a single elementdef minCost(a, i, j, k, prefix, dp): if (i >= j): return 0 # Check if the value is # already stored in the array if (dp[i][j] != -1): return dp[i][j] best_cost = inf for pos in range(i, j): # Compute left subproblem left = minCost(a, i, pos, k, prefix, dp) # Compute left subproblem right = minCost(a, pos + 1, j, k, prefix, dp) # Calculate minimum cost best_cost = min(best_cost, left + right + (k * Combine(prefix, i, j))) # Store the answer to # avoid recalculation dp[i][j] = best_cost return dp[i][j]# Function to generate the cost using# Prefix Sum Array techniquedef preprocess(a, n): p = [0] * n p[0] = a[0] for i in range(1, n): p[i] = p[i - 1] + a[i] return p# Function to combine the sum# of the two subproblemsdef Combine(p, i, j): if (i == 0): return p[j] else: return p[j] - p[i - 1]# Driver Codeif __name__ == "__main__": n = 4 a = [ 4, 5, 6, 7 ] k = 3 # Initialise dp array dp = [[-1 for x in range (n + 1)] for y in range (n + 1)] # Preprocessing the array prefix = preprocess(a, n) print(minCost(a, 0, n - 1, k, prefix, dp))# This code is contributed by chitranayal |
C#
// C# Program for the above approachusing System;class GFG{ static int inf = 10000000; // Function to minimize the cost to // add the array elements to a single element public static int minCost(int []a, int i, int j, int k, int[] prefix, int[,] dp) { if (i >= j) return 0; // Check if the value is // already stored in the array if (dp[i, j] != -1) return dp[i, j]; int best_cost = inf; for (int pos = i; pos < j; pos++) { // Compute left subproblem int left = minCost(a, i, pos, k, prefix, dp); // Compute left subproblem int right = minCost(a, pos + 1, j, k, prefix, dp); // Calculate minimum cost best_cost = Math.Min(best_cost, left + right + (k * Combine(prefix, i, j))); } // Store the answer to // avoid recalculation return dp[i, j] = best_cost; } // Function to generate the cost using // Prefix Sum Array technique public static int[] preprocess(int[] a, int n) { int []p = new int[n]; p[0] = a[0]; for (int i = 1; i < n; i++) p[i] = p[i - 1] + a[i]; return p; } // Function to combine the sum of the two subproblems public static int Combine(int[] p, int i, int j) { if (i == 0) return p[j]; else return p[j] - p[i - 1]; } // Driver Code public static void Main(String []args) { int n = 4; int []a = { 4, 5, 6, 7 }; int k = 3; // Initialise dp array int [,]dp = new int[n + 1, n + 1]; for(int i = 0; i < n + 1; i++) { for (int j = 0; j < n + 1; j++) { dp[i, j] = -1; } } // Preprocessing the array int []prefix = preprocess(a, n); Console.WriteLine(minCost(a, 0, n - 1, k, prefix, dp)); }}// This code is contributed by sapnasingh4991 |
Javascript
<script>// JavaScript program for the above approach var inf = 10000000; // Function to generate the cost using // Prefix Sum Array technique function preprocess(a, n) { var p = Array(n); p[0] = a[0]; for(var i = 1; i < n; i++) { p[i] = p[i - 1] + a[i]; } return p; }// Function to combine the sum of the// two subproblems function Combine(p, i, j) { if (i == 0) return p[j]; else return p[j] - p[i - 1]; } // Function to minimize the cost to // add the array elements to a single element function minCost(a, i, j, k, prefix, dp) { if (i >= j) return 0; // Check if the value is // already stored in the array if (dp[i][j] != -1) return dp[i][j]; var best_cost = inf; for(var pos = i; pos < j; pos++) { // Compute left subproblem var left = minCost(a, i, pos, k, prefix, dp); // Compute left subproblem var right = minCost(a, pos + 1, j, k, prefix, dp); // Calculate minimum cost best_cost = Math.min(best_cost, left + right + (k * Combine(prefix, i, j))); } // Store the answer to // avoid recalculation return dp[i][j] = best_cost; } // Driver code var n = 4; var a = [4, 5, 6, 7]; var k = 3; // Initialise dp array var dp = Array.from(Array(n+1), ()=>Array(n+1).fill(-1));// Preprocessing the array var prefix = preprocess(a, n); document.write( minCost(a, 0, n - 1, k, prefix, dp))</script> |
Output:
132
Time Complexity: O(N2)
Auxiliary Space: O(N2)
Another approach : Using DP Tabulation method ( Iterative approach )
In this approach we use Dp to store computation of subproblems and get the desired output without the help of recursion.
Steps to solve this problem :
- Create a table DP to store the solution of the subproblems .
- Initialize the table with base cases
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in dp[0][n-1].
Implementation :
C++
// C++ code for above approach#include <bits/stdc++.h>using namespace std;int inf = 10000000;// Function to generate the cost using// Prefix Sum Array techniquevector<int> preprocess(vector<int> a, int n){ vector<int> p(n); p[0] = a[0]; for(int i = 1; i < n; i++) { p[i] = p[i - 1] + a[i]; } return p;}// Function to combine the sum of the// two subproblemsint Combine(vector<int> p, int i, int j){ if (i == 0) return p[j]; else return p[j] - p[i - 1];}// Function to minimize the cost to// add the array elements to a single elementint minCost(vector<int> a, int n, int k){ // Initialise dp array vector<vector<int>> dp(n + 1, vector<int>(n + 1)); // Preprocessing the array vector<int> prefix = preprocess(a, n); // Base case for(int i = 0; i < n; i++) { dp[i][i] = 0; } // DP tabulation for(int len = 2; len <= n; len++) { for(int i = 0; i <= n - len; i++) { int j = i + len - 1; dp[i][j] = inf; for(int pos = i; pos < j; pos++) { int left = dp[i][pos]; int right = dp[pos + 1][j]; int cost = k * Combine(prefix, i, j); dp[i][j] = min(dp[i][j], left + right + cost); } } } // return answer return dp[0][n - 1];}// Driver codeint main(){ int n = 4; vector<int> a = { 4, 5, 6, 7 }; int k = 3; // function call cout << minCost(a, n, k) << endl; return 0;}// this code is contributed by bhardwajji |
Java
import java.util.*;public class Main { static int inf = 10000000; // Function to generate the cost using // Prefix Sum Array technique static List<Integer> preprocess(List<Integer> a, int n) { List<Integer> p = new ArrayList<>(); p.add(a.get(0)); for (int i = 1; i < n; i++) { p.add(p.get(i - 1) + a.get(i)); } return p; } // Function to combine the sum of the // two subproblems static int Combine(List<Integer> p, int i, int j) { if (i == 0) return p.get(j); else return p.get(j) - p.get(i - 1); } // Function to minimize the cost to // add the array elements to a single element static int minCost(List<Integer> a, int n, int k) { // Initialise dp array int[][] dp = new int[n + 1][n + 1]; // Preprocessing the array List<Integer> prefix = preprocess(a, n); // Base case for (int i = 0; i < n; i++) { dp[i][i] = 0; } // DP tabulation for (int len = 2; len <= n; len++) { for (int i = 0; i <= n - len; i++) { int j = i + len - 1; dp[i][j] = inf; for (int pos = i; pos < j; pos++) { int left = dp[i][pos]; int right = dp[pos + 1][j]; int cost = k * Combine(prefix, i, j); dp[i][j] = Math.min( dp[i][j], left + right + cost); } } } // return answer return dp[0][n - 1]; } // Driver code public static void main(String[] args) { int n = 4; List<Integer> a = new ArrayList<>(Arrays.asList(4, 5, 6, 7)); int k = 3; // function call System.out.println(minCost(a, n, k)); }} |
Python3
# codeINF = 10000000# Function to generate the cost using# Prefix Sum Array techniquedef preprocess(a, n): p = [0] * n p[0] = a[0] for i in range(1, n): p[i] = p[i - 1] + a[i] return p# Function to combine the sum of the# two subproblemsdef Combine(p, i, j): if i == 0: return p[j] else: return p[j] - p[i - 1]# Function to minimize the cost to# add the array elements to a single elementdef minCost(a, n, k): # Initialise dp array dp = [[0] * (n + 1) for i in range(n + 1)] # Preprocessing the array prefix = preprocess(a, n) # Base case for i in range(n): dp[i][i] = 0 # DP tabulation for length in range(2, n + 1): for i in range(n - length + 1): j = i + length - 1 dp[i][j] = INF for pos in range(i, j): left = dp[i][pos] right = dp[pos + 1][j] cost = k * Combine(prefix, i, j) dp[i][j] = min(dp[i][j], left + right + cost) # return answer return dp[0][n - 1]if __name__ == '__main__': n = 4 a = [4, 5, 6, 7] k = 3 # function call print(minCost(a, n, k)) |
C#
using System;using System.Collections.Generic;class Program{ static int inf = 10000000; // Function to generate the cost using // Prefix Sum Array technique static List<int> preprocess(List<int> a, int n) { List<int> p = new List<int>(); p.Add(a[0]); for (int i = 1; i < n; i++) { p.Add(p[i - 1] + a[i]); } return p; } // Function to combine the sum of the // two subproblems static int Combine(List<int> p, int i, int j) { if (i == 0) return p[j]; else return p[j] - p[i - 1]; } // Function to minimize the cost to // add the array elements to a single element static int minCost(List<int> a, int n, int k) { // Initialise dp array int[,] dp = new int[n + 1, n + 1]; // Preprocessing the array List<int> prefix = preprocess(a, n); // Base case for (int i = 0; i < n; i++) { dp[i, i] = 0; } // DP tabulation for (int len = 2; len <= n; len++) { for (int i = 0; i <= n - len; i++) { int j = i + len - 1; dp[i, j] = inf; for (int pos = i; pos < j; pos++) { int left = dp[i, pos]; int right = dp[pos + 1, j]; int cost = k * Combine(prefix, i, j); dp[i, j] = Math.Min(dp[i, j], left + right + cost); } } } // return answer return dp[0, n - 1]; } // Driver code static void Main(string[] args) { int n = 4; List<int> a = new List<int>() { 4, 5, 6, 7 }; int k = 3; // function call Console.WriteLine(minCost(a, n, k)); }} |
Javascript
// JavaScript code for above approachconst inf = 10000000;// Function to generate the cost using// Prefix Sum Array techniquefunction preprocess(a, n) { let p = []; p[0] = a[0]; for (let i = 1; i < n; i++) { p[i] = p[i - 1] + a[i]; } return p;}// Function to combine the sum of the// two subproblemsfunction Combine(p, i, j) { if (i == 0) return p[j]; else return p[j] - p[i - 1];}// Function to minimize the cost to// add the array elements to a single elementfunction minCost(a, n, k) { // Initialise dp array let dp = new Array(n + 1).fill().map(() => new Array(n + 1).fill(0)); // Preprocessing the array let prefix = preprocess(a, n); // Base case for (let i = 0; i < n; i++) { dp[i][i] = 0; } // DP tabulation for (let len = 2; len <= n; len++) { for (let i = 0; i <= n - len; i++) { let j = i + len - 1; dp[i][j] = inf; for (let pos = i; pos < j; pos++) { let left = dp[i][pos]; let right = dp[pos + 1][j]; let cost = k * Combine(prefix, i, j); dp[i][j] = Math.min(dp[i][j], left + right + cost); } } } // return answer return dp[0][n - 1];}// Driver codelet n = 4;let a = [4, 5, 6, 7];let k = 3;// function callconsole.log(minCost(a, n, k));// This code is contributed by user_dtewbxkn77n |
Output
132
Time Complexity: O(N^3)
Auxiliary Space: O(N^2)
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