Rearrange given array to obtain positive prefix sums at exactly X indices
Last Updated :
15 Sep, 2021
Given an array arr[] consisting of N integers whose absolute values are distinct and an integer K, the task is to find such an arrangement of the given array by the following operations, such that it has positive prefix sum at exactly K places:
- Choose two integers i, j and swap the values of the element at indices i and j, where 0 ? i, j < N and i ? j.
- Choose an index i and change its sign i.e., change positive to negative and vice-versa where 0 ? i < N.
Examples:
Input: arr[] = {1, -2, 4, 5, -3}, K = 3
Output: {-1, 2, -4, 3, 5}
Explanation:
Prefix sums of the given array are {-1, 1, -3, 0, 5}.
Therefore, exactly 3 indices have positive prefix sum.
Input: arr[] = {1, 4, 3, 2}, K = 2
Output: {1, -4, 2, 3}
Explanation:
Prefix sums of the given array are 1, -3, -1, 2}.
Therefore, exactly 2 indices have positive prefix sum.
Approach: The given problem can be solved by observing that one can get the sorted array using only the first type of operation. And one can change every element into positive elements using the 2nd type of operation. Follow the steps below to solve the problem:
- Make all the values of the array positive by using the 2nd operation i.e., flip the sign of all negative numbers to positive.
- Sort the array arr[] in ascending order and set K equal to (N – K).
- If the sum of K and the count of 0s in the given array arr[] is greater than N, then there can’t be any such possible arrangement. Hence print “-1”. Else perform the below operations.
- Traverse the array over the range [0, N – 1] and follow the below steps:
- Change the ith element into negative by the 2nd operation.
- Increment i by 2 and decrement K by 1.
- If K is equal to 0 then break.
- Check if K is still greater than 0, then traverse the array from last until K is greater than 0, and change every positive element into negative and decrement K by 1.
- After the above steps, print the array as the required arrangement.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void rearrange(int* a, int n, int x)
{
for (int i = 0; i < n; i++)
a[i] = abs(a[i]);
sort(a, a + n);
x = n - x;
int z = count(a, a + n, 0);
if (x > n - z) {
cout << "-1\n";
return;
}
for (int i = 0; i < n && x > 0;
i += 2) {
a[i] = -a[i];
x--;
}
for (int i = n - 1; i >= 0
&& x > 0;
i--) {
if (a[i] > 0) {
a[i] = -a[i];
x--;
}
}
for (int i = 0; i < n; i++) {
cout << a[i] << " ";
}
}
int main()
{
int arr[] = { 0, -2, 4, 5, -3 };
int K = 3;
int N = sizeof(arr) / sizeof(arr[0]);
rearrange(arr, N, K);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int count(int[] a)
{
int counter = 0;
for(int i = 0; i < a.length; i++)
{
if (a[i] == 0)
{
counter++;
}
}
return counter;
}
static void rearrange(int[] a, int n, int x)
{
for(int i = 0; i < n; i++)
a[i] = Math.abs(a[i]);
Arrays.sort(a);
x = n - x;
int z = count(a);
if (x > n - z)
{
System.out.println("-1");
return;
}
for(int i = 0; i < n && x > 0; i += 2)
{
a[i] = -a[i];
x--;
}
for(int i = n - 1; i >= 0 && x > 0; i--)
{
if (a[i] > 0)
{
a[i] = -a[i];
x--;
}
}
for(int i = 0; i < n; i++)
{
System.out.print(a[i] + " ");
}
}
public static void main (String[] args)
{
int arr[] = { 0, -2, 4, 5, -3 };
int K = 3;
int N = arr.length;
rearrange(arr, N, K);
}
}
|
Python3
def rearrange(a, n, x):
for i in range(n):
a[i] = abs(a[i])
a = sorted(a)
x = n - x;
z = a.count(0)
if (x > n - z):
print("-1")
return
for i in range(0, n, 2):
if x <= 0:
break
a[i] = -a[i]
x -= 1
for i in range(n - 1, -1, -1):
if x <= 0:
break
if (a[i] > 0):
a[i] = -a[i]
x -= 1
for i in range(n):
print(a[i], end = " ")
if __name__ == '__main__':
arr = [0, -2, 4, 5, -3]
K = 3
N = len(arr)
rearrange(arr, N, K)
|
C#
using System;
class GFG{
static int count(int[] a)
{
int counter = 0;
for(int i = 0; i < a.Length; i++)
{
if (a[i] == 0)
{
counter++;
}
}
return counter;
}
static void rearrange(int[] a, int n, int x)
{
for(int i = 0; i < n; i++)
a[i] = Math.Abs(a[i]);
Array.Sort(a);
x = n - x;
int z = count(a);
if (x > n - z)
{
Console.WriteLine("-1");
return;
}
for(int i = 0; i < n && x > 0; i += 2)
{
a[i] = -a[i];
x--;
}
for(int i = n - 1; i >= 0 && x > 0; i--)
{
if (a[i] > 0)
{
a[i] = -a[i];
x--;
}
}
for(int i = 0; i < n; i++)
{
Console.Write(a[i] + " ");
}
}
public static void Main(String[] args)
{
int []arr = { 0, -2, 4, 5, -3 };
int K = 3;
int N = arr.Length;
rearrange(arr, N, K);
}
}
|
Javascript
<script>
function count(a)
{
let counter = 0;
for(let i = 0; i < a.length; i++)
{
if (a[i] == 0)
{
counter++;
}
}
return counter;
}
function rearrange(a, n, x)
{
for(let i = 0; i < n; i++)
a[i] = Math.abs(a[i]);
a.sort();
x = n - x;
let z = count(a);
if (x > n - z)
{
document.write("-1");
return;
}
for(let i = 0; i < n && x > 0; i += 2)
{
a[i] = -a[i];
x--;
}
for(let i = n - 1; i >= 0 && x > 0; i--)
{
if (a[i] > 0)
{
a[i] = -a[i];
x--;
}
}
for(let i = 0; i < n; i++)
{
document.write(a[i] + " ");
}
}
let arr = [ 0, -2, 4, 5, -3 ];
let K = 3;
let N = arr.length;
rearrange(arr, N, K);
</script>
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Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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