Minimum removals from array to make max â€“ min <= K in C++

Problem statement

Given N integers and K, find the minimum number of elements that should be removed such that Amax - Amin <= K. After removal of elements, Amax and Amin is considered among the remaining elements

Examples

If arr[] = {1, 3, 4, 9, 10, 11, 12, 17, 20} and k = 4 then output would be 5:

• Remove 1, 3 and 4 from beginning of an array
• Remove 17 and 20 from the end of array
• Final array becomes {9, 10, 11, 12} where 12 – 9 <= 4

Algorithm

1. Sort the given elements
2. Using greedy approach, the best way is to remove the minimum element or the maximum element and then check if Amax - Amin <= K. There are various combinations of removals that have to be considered.
3. There will be two possible ways of removal, either we remove the minimum or we remove the maximum. Let(i…j) be the index of elements left after removal of elements. Initially, we start with i=0 and j=n-1 and the number of elements removed is 0 at the beginnings.
4. We only remove an element if a[j]-a[i]>k, the two possible ways of removal are (i+1…j) or (i…j-1). The minimum of the two is considered

Example

Live Demo

#include <bits/stdc++.h>
#define MAX 100
using namespace std;
int dp[MAX][MAX];
int removeCombinations(int *arr, int i, int j, int k) {
if (i >= j) {
return 0;
} else if ((arr[j] - arr[i]) <= k) {
return 0;
} else if (dp[i][j] != -1) {
return dp[i][j];
} else if ((arr[j] - arr[i]) > k) {
dp[i][j] = 1 + min(removeCombinations(arr, i +
1, j, k),
removeCombinations(arr, i, j - 1,k));
}
return dp[i][j];
}
int removeNumbers(int *arr, int n, int k){
sort(arr, arr + n);
memset(dp, -1, sizeof(dp));
return n == 1 ? 0 : removeCombinations(arr, 0, n - 1,k);
}
int main() {
int arr[] = {1, 3, 4, 9, 10, 11, 12, 17, 20};
int n = sizeof(arr) / sizeof(arr[0]);
int k = 4;
cout << "Minimum numbers to be removed = " <<
removeNumbers(arr, n, k) << endl;
return 0;
}

When you compile and execute above program. It generates following output

Output

Minimum numbers to be removed = 5