710035. Array Removals
Array Removals
Slug: array-removals
Difficulty: Medium
Id: 710035
Topic Tags: Arrays, Dynamic Programming, Data Structures, Algorithms
Company Tags: None
Summary
Given an array arr of integers and two indices i and j, the problem is to find the minimum number of elements that need to be removed from the array such that all the elements in the range [i, j] are removed. The goal is to minimize the total number of removals while ensuring that all elements in the specified range are removed.
Detailed Explanation
The problem can be solved by using dynamic programming. We can maintain a 2D table dp[i][j] where dp[i][j] represents the minimum number of removals needed to remove all elements in the range [i, j]. The base case for this recursion is when i and j are equal, which means we need to remove only one element. In this case, the answer is 1.
For other cases, we can consider two options: either include the current element in our removals or exclude it. If we include the current element, we would have to remove all elements up to i-1 (since we cannot leave any gaps) and then remove all elements from j+1 onwards. This would give us a total of dp[i-1][j] + j-i + 2.
On the other hand, if we exclude the current element, we can simply use the minimum number of removals needed to remove all elements in the range [i, j-1]. This gives us dp[i][j-1].
We then choose the option that results in a smaller total number of removals and add it to our answer. The final answer is stored in dp[0][n-1], where n is the length of the array.
Here's a step-by-step breakdown of the solution:
- Initialize the 2D table
dp[i][j]with all elements set to infinity. - For each element
arr[k]in the range[i, j], calculate the minimum number of removals needed by considering two options: including or excluding the current element. - Choose the option that results in a smaller total number of removals and add it to our answer.
- Store the final answer in
dp[0][n-1].
Time complexity: O(n^2), where n is the length of the array.
Space complexity: O(n^2) for storing the 2D table dp[i][j].
Optimized Solutions
Java
public int minRemovals(int[] arr, int i, int j) {
int n = arr.length;
int[][] dp = new int[n][n];
for (int k = 0; k < n; k++) {
for (int l = 0; l < n; l++) {
if (k == l) {
dp[k][l] = 1;
} else if (k > l) {
dp[k][l] = Integer.MAX_VALUE;
} else {
if (arr[l] >= i && arr[l] <= j) {
dp[k][l] = 1 + dp[k-1][l-1];
} else {
dp[k][l] = Math.min(dp[k-1][l], dp[k][l-1]) + 2;
}
}
}
}
return dp[0][n-1];
}