Minimize Cost to reduce the Array to a single element by given operations
Last Updated :
19 Apr, 2023
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 halves
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;
}
// Function to minimize the cost to
// reduce the array to a single element
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 = min(best_cost, left + right +
k * Combine(a, i, j));
}
// Return the answer
return best_cost;
}
// Driver code
int 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 Program to implement
// the above approach
import 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 Program to implement
# the above approach
inf = 10000009;
# Function to minimize the cost to
# reduce the array to a single element
def 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 halves
def 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 code
if __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# Program to implement
// the above approach
using 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
<script>
// JavaScript Program to implement
// the above approach
var inf = 10000009;
// Function to combine the sum of the two halves
function 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 element
function 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 code
var n = 4;
var a = [4, 5, 6, 7];
var k = 3;
document.write( minCost(a, 0, n - 1, k));
</script>
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 Program for the above approach
import 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 program for the above approach
inf = 10000000
# Function to minimize the cost to
# add the array elements to a single element
def 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 technique
def 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 subproblems
def Combine(p, i, j):
if (i == 0):
return p[j]
else:
return p[j] - p[i - 1]
# Driver Code
if __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# Program for the above approach
using 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
<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>
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 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 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 code
int 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
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));
}
}
# code
INF = 10000000
# Function to generate the cost using
# Prefix Sum Array technique
def 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 subproblems
def 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
def 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))
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 code for above approach
const inf = 10000000;
// Function to generate the cost using
// Prefix Sum Array technique
function 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 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, 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 code
let n = 4;
let a = [4, 5, 6, 7];
let k = 3;
// function call
console.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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