N consecutive ropes problem
Last Updated :
21 Dec, 2022
Given N ropes, each rope has a length associated with it. At a time, only two consecutive small ropes tied at end form a large rope and cost of forming sum of their length. Find the minimum cost when all ropes are tied to form a single rope.
Examples:
Input: arr[] = {7, 6, 8, 6, 1, 1}
Output: 68
{7, 6, 8, 6, 1, 1} – {7, 6, 8, 6, 2}, cost = 2
{7, 6, 8, 6, 2} – {7, 6, 8, 8}, cost = 8
{7, 6, 8, 8} – {13, 8, 8}, cost = 13
{13, 8, 8} – {13, 16}, cost = 16
{13, 16} – {29}, cost = 29
2 + 8 + 13 + 16 + 29 = 68
Input: arr[] = {10, 20, 30, 40}
Output: 190
Approach: A similar problem has been discussed in this article where any two ropes can be connected but in this problem, only the consecutive ropes can be connected. This problem can be solved by matrix chain multiplication technique of dynamic programming.
Optimal Substructure: In the 2nd example, the ropes can be connected as (((10 + 20) + 30) + 40)
Connect 10 and 20; cost = 30; expression becomes ((30 + 30) + 40)
Connect 30 and 30; cost = 90; expression becomes (60 + 40)
Connect 60 and 40; cost = 190 and we get a single rope
A simple solution is to place parenthesis at all possible places then calculate the cost for each segment and sum up to total cost. This can be done for all the valid parenthesis sequence and the minimum will be the answer.
Given ropes of length r1, r2, r3 and r4.
The connection can be formed in the following ways:
(((r1 + r2) + r3) + r4) or
((r1 + (r2 + r3)) + r4) or
((r1 + r2) + (r3 + r4)) …
Total cost of the 1st way is = r4 + 2 * r3 + 3 * (r2 + r1)
So when the set of parenthesis are placed, the problem is divided into subproblems of smaller sizes. Therefore, the problem has optimal substructure property and can be easily solved using recursion.
Overlapping Subproblems
(((r1 + r2) + r3) + r4)
and
((r1 + r2) + (r3 + r4))
both has a common part i.e. (r1 + r2)
So the solution is as follows:
- Precompute sum of different interval to save computation time.
- If we want two segments of {arr[i] arr[i+1] …arr[k]} and {arr[k+1] arr[k+2] …arr[j]} to connect, then our cost will be
MinCost(i, k) + MinCost(k + 1, j) + sum(arr[i] to arr[j])
- where MinCost(i, k) = minimum cost in range(i, k) and sum(arr[i] to arr[j]) = cost to connect the rope segment of (i, k) and (k + 1, j). We can store the subproblems in dp table to save computation time.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int MinCost(int arr[], int n)
{
int dp[n + 5][n + 5], sum[n + 5][n + 5];
memset(sum, 0, sizeof(0));
for (int i = 0; i < n; i++) {
int k = arr[i];
for (int j = i; j < n; j++) {
if (i == j)
sum[i][j] = k;
else {
k += arr[j];
sum[i][j] = k;
}
}
}
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
dp[i][j] = INT_MAX;
if (i == j)
dp[i][j] = 0;
else {
for (int k = i; k < j; k++) {
dp[i][j] = min(dp[i][j],
dp[i][k] + dp[k + 1][j]
+ sum[i][j]);
}
}
}
}
return dp[0][n - 1];
}
int main()
{
int arr[] = { 7, 6, 8, 6, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << MinCost(arr, n);
return 0;
}
|
Java
class GFG
{
static int MinCost(int arr[], int n)
{
int [][]dp = new int[n + 5][n + 5];
int [][]sum = new int[n + 5][n + 5];
for (int i = 0; i < n; i++)
{
int k = arr[i];
for (int j = i; j < n; j++)
{
if (i == j)
sum[i][j] = k;
else
{
k += arr[j];
sum[i][j] = k;
}
}
}
for (int i = n - 1; i >= 0; i--)
{
for (int j = i; j < n; j++)
{
dp[i][j] = Integer.MAX_VALUE;
if (i == j)
dp[i][j] = 0;
else
{
for (int k = i; k < j; k++)
{
dp[i][j] = Math.min(dp[i][j],
dp[i][k] +
dp[k + 1][j] +
sum[i][j]);
}
}
}
}
return dp[0][n - 1];
}
public static void main(String []args)
{
int arr[] = { 7, 6, 8, 6, 1, 1 };
int n = arr.length;
System.out.println(MinCost(arr, n));
}
}
|
Python3
def MinCost(arr, n):
dp = [[0 for i in range(n + 5)]
for i in range(n + 5)]
sum = [[0 for i in range(n + 5)]
for i in range(n + 5)]
for i in range(n):
k = arr[i]
for j in range(i, n):
if (i == j):
sum[i][j] = k
else:
k += arr[j]
sum[i][j] = k
for i in range(n - 1, -1, -1):
for j in range(i, n):
dp[i][j] = 10**9
if (i == j):
dp[i][j] = 0
else :
for k in range(i, j):
dp[i][j] = min(dp[i][j], dp[i][k] +
dp[k + 1][j] + sum[i][j])
return dp[0][n - 1]
arr = [7, 6, 8, 6, 1, 1]
n = len(arr)
print(MinCost(arr, n))
|
C#
using System;
class GFG
{
static int MinCost(int []arr, int n)
{
int [,]dp = new int[n + 5, n + 5];
int [,]sum = new int[n + 5, n + 5];
for (int i = 0; i < n; i++)
{
int k = arr[i];
for (int j = i; j < n; j++)
{
if (i == j)
sum[i, j] = k;
else
{
k += arr[j];
sum[i, j] = k;
}
}
}
for (int i = n - 1; i >= 0; i--)
{
for (int j = i; j < n; j++)
{
dp[i, j] = int.MaxValue;
if (i == j)
dp[i, j] = 0;
else
{
for (int k = i; k < j; k++)
{
dp[i, j] = Math.Min(dp[i, j],
dp[i, k] +
dp[k + 1, j] +
sum[i, j]);
}
}
}
}
return dp[0, n - 1];
}
public static void Main(String []args)
{
int []arr = { 7, 6, 8, 6, 1, 1 };
int n = arr.Length;
Console.WriteLine(MinCost(arr, n));
}
}
|
Javascript
<script>
function MinCost(arr, n) {
let dp = new Array(n + 5);
let sum = new Array(n + 5);
for (let i = 0; i < n + 5; i++) {
dp[i] = [];
sum[i] = [];
for (let j = 0; j < n + 5; j++) {
dp[i].push(0)
sum[i].push(0)
}
}
console.log(dp)
for (let i = 0; i < n; i++)
{
let k = arr[i];
for (let j = i; j < n; j++)
{
if (i == j)
sum[i][j] = k;
else
{
k += arr[j];
sum[i][j] = k;
}
}
}
for (let i = n - 1; i >= 0; i--)
{
for (let j = i; j < n; j++)
{
dp[i][j] = Number.MAX_SAFE_INTEGER;
if (i == j)
dp[i][j] = 0;
else
{
for (let k = i; k < j; k++)
{
dp[i][j] = Math.min(dp[i][j], dp[i][k] +
dp[k + 1][j] + sum[i][j]);
}
}
}
}
return dp[0][n - 1];
}
let arr = [7, 6, 8, 6, 1, 1];
let n = arr.length;
document.write(MinCost(arr, n));
</script>
|
Time Complexity: O(n3)
Auxiliary Space: O(n2), where n is the size of the given array.
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