We have been given N balloons, each with a number of coins associated with it. On bursting a balloon i, the number of coins gained is equal to A[i-1]*A[i]*A[i+1]. Also, balloons i-1 and i+1 now become adjacent. Find the maximum possible profit earned after bursting all the balloons. Assume an extra 1 at each boundary.
Examples:
Input : 5, 10
Output : 60
Explanation - First Burst 5, Coins = 1*5*10
Then burst 10, Coins+= 1*10*1
Total = 60
Input : 1, 2, 3, 4, 5
Output : 110
A recursive solution is discussed here. We can solve this problem using dynamic programming.
First, consider a sub-array from indices Left to Right(inclusive).
If we assume the balloon at index Last to be the last balloon to be burst in this sub-array, we would say the coined gained to be-A[left-1]*A[last]*A[right+1].
Also, the total Coin Gained would be this value, plus dp[left][last – 1] + dp[last + 1][right], where dp[i][j] means maximum coin gained for sub-array with indices i, j.
Therefore, for each value of Left and Right, we need find and choose a value of Last with maximum coin gained, and update the dp array.
Our Answer is the value at dp[1][N].
C++
// C++ program burst balloon problem#include <bits/stdc++.h>#include <iostream>using namespace std;int getMax(int A[], int N){ // Add Bordering Balloons int B[N + 2]; B[0] = 1; B[N + 1] = 1; for (int i = 1; i <= N; i++) B[i] = A[i - 1]; // Declare DP Array int dp[N + 2][N + 2]; memset(dp, 0, sizeof(dp)); for (int length = 1; length < N + 1; length++) { for (int left = 1; left < N - length + 2; left++) { int right = left + length - 1; // For a sub-array from indices left, right // This innermost loop finds the last balloon burst for (int last = left; last < right + 1; last++) { dp[left][right] = max(dp[left][right], dp[left][last - 1] + B[left - 1] * B[last] * B[right + 1] + dp[last + 1][right]); } } } return dp[1][N];}// Driver code int main(){ int A[] = { 1, 2, 3, 4, 5 }; // Size of the array int N = sizeof(A) / sizeof(A[0]); // Calling function cout << getMax(A, N) << endl;}// This code is contributed by ashutosh450 |
Java
// Java program to illustrate// Burst balloon problemimport java.util.Arrays;class GFG{public static int getMax(int[] A, int N) { // Add Bordering Balloons int[] B = new int[N + 2]; B[0] = B[N + 1] = 1; for(int i = 1; i <= N; i++) B[i] = A[i - 1]; // Declaring DP array int[][] dp = new int[N + 2][N + 2]; for(int length = 1; length < N + 1; length++) { for(int left = 1; left < N - length + 2; left++) { int right = left + length -1; // For a sub-array from indices // left, right. This innermost // loop finds the last balloon burst for(int last = left; last < right + 1; last++) { dp[left][right] = Math.max( dp[left][right], dp[left][last - 1] + B[left - 1] * B[last] * B[right + 1] + dp[last + 1][right]); } } } return dp[1][N];}// Driver codepublic static void main(String args[]){ int[] A = { 1, 2, 3, 4, 5 }; // Size of the array int N = A.length; // Calling function System.out.println(getMax(A, N)); }}// This code is contributed by dadi madhav |
Python3
# Python3 program burst balloon problem. def getMax(A): N = len(A) A = [1] + A + [1]# Add Bordering Balloons dp = [[0 for x in range(N + 2)] for y in range(N + 2)]# Declare DP Array for length in range(1, N + 1): for left in range(1, N-length + 2): right = left + length -1 # For a sub-array from indices left, right # This innermost loop finds the last balloon burst for last in range(left, right + 1): dp[left][right] = max(dp[left][right], \ dp[left][last-1] + \ A[left-1]*A[last]*A[right + 1] + \ dp[last + 1][right]) return(dp[1][N])# Driver codeA = [1, 2, 3, 4, 5]print(getMax(A)) |
C#
// C# program to illustrate // Burst balloon problemusing System; class GFG{ public static int getMax(int[] A, int N) { // Add Bordering Balloons int[] B = new int[N + 2]; B[0] = B[N + 1] = 1; for(int i = 1; i <= N; i++) B[i] = A[i - 1]; // Declaring DP array int[,] dp = new int[(N + 2), (N + 2)]; for(int length = 1; length < N + 1; length++) { for(int left = 1; left < N - length + 2; left++) { int right = left + length -1; // For a sub-array from indices // left, right. This innermost // loop finds the last balloon burst for(int last = left; last < right + 1; last++) { dp[left, right] = Math.Max( dp[left, right], dp[left, last - 1] + B[left - 1] * B[last] * B[right + 1] + dp[last + 1, right]); } } } return dp[1, N]; } // Driver code public static void Main() { int[] A = new int[] { 1, 2, 3, 4, 5 }; // Size of the array int N = A.Length; // Calling function Console.WriteLine(getMax(A, N)); } }// This code is contributed by sanjoy_62 |
Javascript
<script>// Javascript program burst balloon problemfunction getMax(A, N){ // Add Bordering Balloons var B = new Array(N+2); B[0] = 1; B[N + 1] = 1; for (var i = 1; i <= N; i++) B[i] = A[i - 1]; // Declare DP Array var dp = new Array(N + 2); for (var i = 0; i < dp.length; i++) { dp[i] = new Array(N + 2).fill(0); } for (var length = 1; length < N + 1; length++) { for (var left = 1; left < N - length + 2; left++) { var right = left + length - 1; // For a sub-array from indices left, right // This innermost loop finds the last balloon burst for (var last = left; last < right + 1; last++) { dp[left][right] = Math.max(dp[left][right], dp[left][last - 1] + B[left - 1] * B[last] * B[right + 1] + dp[last + 1][right]); } } } return dp[1][N];}// Driver code var A = [ 1, 2, 3, 4, 5 ];// Size of the arrayvar N = A.length;// Calling functiondocument.write(getMax(A, N));// This code is contributed by shubhamsingh10</script> |
Output:
110
Time Complexity: O(N3)
Auxiliary Space: O(N2)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
