Minimum deletions to be done in given array such that every pair sum is a power of 2
Last Updated :
29 Dec, 2022
Given an array arr[] consisting of N integers the task is to find the minimum number of elements that should be removed, such that for every remaining element arr[i], there exists another element arr[j], (i!=j) such that the sum of arr[i] and arr[j] is a power of 2. If after any number of deletions, no such pair exists, print -1.
Examples:
Input: arr[] ={1, 2, 3, 4, 5, 6}, N = 6
Output: 1
Explanation:
Removing 4 is sufficient as, for the remaining elements there exist an element such that their sum is a power of 2:
- For 1, there exists 3 such that (1+3=4) is a power of 2.
- For 2, there exists 6 such that (2+6=8) is a power of 2.
- For 3, there exists 1 such that (3+1=4) is a power of 2.
- For 5, there exists 3 such that (5+3=8) is a power of 2.
- For 6, there exists 2 such that (6+2=8) is a power of 2.
Input: A={1, 5, 10, 25, 50}, N=5
Output: -1
Approach: The idea is to use bitwise operators and the map to keep track of the frequencies of every element. Follow the steps below to solve the problem:
- Initialize a variable ans to 0, to store the number of elements to be removed.
- Calculate the frequency of all elements in arr[], and store them in a map say mp.
- Iterate from 0 to N-1, and for each current index i, do the following:
- Initialize a variable say flag to 0.
- Iterate from 0 to 30, and for each current index j, do the following:
- If the difference between 2j and arr[i] is equal to arr[i], and the frequency of arr[i] is greater than 1, increment flag and break.
- Otherwise, if the frequency of the difference between 2j and arr[i] is greater than 0, increment flag and break.
- If the flag is still 0, increment ans, as the current element needs to be removed.
- Finally, after completing the above steps, print the count of the removed element as the value obtained in ans, if it is less than equal to N. Else print -1.
Below is an implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int minimumDeletions(int arr[], int N)
{
int ans = 0;
map<int, int> mp;
for (int i = 0; i < N; i++)
mp[arr[i]]++;
for (int i = 0; i < N; i++) {
int flag = 0;
for (int j = 0; j < 31; j++) {
int pw = (1 << j);
if (pw - arr[i] == arr[i]) {
if (mp[arr[i]] > 1) {
flag = 1;
break;
}
}
else if (mp[pw - arr[i]] > 0) {
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
return ans == N ? -1 : ans;
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << minimumDeletions(arr, N) << endl;
int arr1[] = { 1, 5, 10, 25, 50 };
N = sizeof(arr) / sizeof(arr[0]);
cout << minimumDeletions(arr1, N) << endl;
return 0;
}
|
Java
import java.util.*;
class GFG{
static int minimumDeletions(int []arr, int N)
{
int ans = 0;
Map<Integer, Integer> mp = new HashMap<Integer, Integer>();
for (int i = 0; i < N; i++){
if(mp.containsKey(arr[i]))
mp.put(arr[i],mp.get(arr[i]) + 1);
else
mp.put(arr[i],1);
}
for (int i = 0; i < N; i++)
{
int flag = 0;
for (int j = 0; j < 31; j++) {
int pw = (1 << j);
if (pw - arr[i] == arr[i]) {
if (mp.containsKey(arr[i]) && mp.get(arr[i]) > 1) {
flag = 1;
break;
}
}
else if (mp.containsKey(pw - arr[i]) && mp.get(pw - arr[i]) > 0) {
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
if(ans == N)
return -1;
else
return ans;
}
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = arr.length;
System.out.println(minimumDeletions(arr, N));
int arr1[] = { 1, 5, 10, 25, 50 };
N = arr1.length;
System.out.print(minimumDeletions(arr1, N));
}
}
|
Python3
def minimumDeletions(arr, N):
ans = 0
mp = {}
for i in arr:
mp[i] = mp.get(i,0)+1
for i in range(N):
flag = 0
for j in range(31):
pw = (1 << j)
if i >= len(arr):
break
if (pw - arr[i] == arr[i]):
if (mp[arr[i]] > 1):
flag = 1
break
elif (((pw - arr[i]) in mp) and mp[pw - arr[i]] > 0):
flag = 1
break
if (flag == 0):
ans += 1
return -1 if ans == N else ans
if __name__ == '__main__':
arr= [1, 2, 3, 4, 5, 6]
N = len(arr)
print (minimumDeletions(arr, N))
arr1= [1, 5, 10, 25, 50]
N = len(arr)
print (minimumDeletions(arr1, N))
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int minimumDeletions(int []arr, int N)
{
int ans = 0;
Dictionary<int,int> mp = new Dictionary<int,int>();
for (int i = 0; i < N; i++){
if(mp.ContainsKey(arr[i]))
mp[arr[i]] += 1;
else
mp.Add(arr[i],1);
}
for (int i = 0; i < N; i++)
{
int flag = 0;
for (int j = 0; j < 31; j++) {
int pw = (1 << j);
if (pw - arr[i] == arr[i]) {
if (mp.ContainsKey(arr[i]) && mp[arr[i]] > 1) {
flag = 1;
break;
}
}
else if (mp.ContainsKey(pw - arr[i]) && mp[pw - arr[i]] > 0) {
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
if(ans == N)
return -1;
else
return ans;
}
public static void Main()
{
int []arr = { 1, 2, 3, 4, 5, 6 };
int N = arr.Length;
Console.WriteLine(minimumDeletions(arr, N));
int []arr1 = { 1, 5, 10, 25, 50 };
N = arr1.Length;
Console.Write(minimumDeletions(arr1, N));
}
}
|
Javascript
<script>
function minimumDeletions(arr, N) {
let ans = 0;
let mp = new Map();
for (let i = 0; i < N; i++) {
if (mp.has(arr[i])) {
mp.set(arr[i], mp.get(arr[i]) + 1)
} else {
mp.set(arr[i], 1)
}
}
for (let i = 0; i < N; i++) {
let flag = 0;
for (let j = 0; j < 31; j++) {
let pw = (1 << j);
if (pw - arr[i] == arr[i]) {
if (mp.get(arr[i]) > 1) {
flag = 1;
break;
}
}
else if (mp.get(pw - arr[i]) > 0) {
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
return ans == N ? -1 : ans;
}
let arr = [1, 2, 3, 4, 5, 6];
let N = arr.length;
document.write(minimumDeletions(arr, N) + "<br>");
let arr1 = [1, 5, 10, 25, 50];
N = arr.length;
document.write(minimumDeletions(arr1, N) + "<br>");
</script>
|
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...