# Minimum number of elements that should be removed to make the array good using C++.

## Problem statement

Given an array “arr”, the task is to find the minimum number of elements to be removed to make the array good.

A sequence a1, a2, a3. . .an is called good if for each element a[i], there exists an element a[j] (i not equals to j) such that a[i] + a[j] is a power of two.

arr1[] = {1, 1, 7, 1, 5}

In above array if we delete element ‘5’ then array becomes good array. After this any pair of arr[i] + arr[j] is power of two −

• arr[0] + arr[1] = (1 + 1) = 2 which power of two
• arr[0] + arr[2] = (1 + 7) = 8 which is power of two

## Algorithm

1. We have to delete only such a[i] for which there is no a[j] such that a[i] + a[i] is a power of 2.
2. For each value find the number of its occurrences in the array
3. Check that a[i] doesn’t have a pair a[j]

## Example

#include <iostream>
#include <map>
#define SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
using namespace std;
int minDeleteRequred(int *arr, int n){
map<int, int> frequency;
for (int i = 0; i < n; ++i) {
frequency[arr[i]]++;
}
int delCnt = 0;
for (int i = 0; i < n; ++i) {
bool doNotRemove = false;
for (int j = 0; j < 31; ++j) {
int pair = (1 << j) - arr[i];
if (frequency.count(pair) &&
(frequency[pair] > 1 ||
(frequency[pair] == 1 &&
pair != arr[i]))) {
doNotRemove = true;
break;
}
}
if (!doNotRemove) {
++delCnt;
}
}
return delCnt;
}
int main(){
int arr[] = {1, 1, 7, 1, 5};
cout << "Minimum elements to be deleted = " << minDeleteRequred(arr, SIZE(arr)) << endl;
return 0;
}

## Output

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

Minimum elements to be deleted = 1

Updated on: 31-Oct-2019

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