Move negatives to the end with relative order preserved using extra space
Last Updated :
23 Jul, 2025
Given an unsorted array arr[] of integers. The task is place all negative element at the end of array such that the relative ordering of all negative elements must remain the same before and after placement, and the same should hold true for all positive elements.
Examples:
Input: arr[] = [1, -1, 3, 2, -7, -5, 11, 6]
Output: [1, 3, 2, 11, 6, -1, -7, -5]
Explanation: All negative elements [-1, -7, -5] were arranged after positive elements [1, 3, 2, 11, 6] and the relative ordering was also preserved.
Input: arr[] = [-5, 7, -3, -4, 9, 10, -1, 11]
Output: [7, 9, 10, 11, -5, -3, -4, -1]
Explanation: All negative elements [-5, -3, -4, -1] were arranged after positive elements [7, 9, 10, 11] and the relative ordering was also preserved.
We have discussed different approaches to this problem in below post.
Rearrange positive and negative numbers with constant extra space
Using Naive Partitioning Algorithm
The idea is to create an auxiliary array (temp[]). First, we traverse the given array and store all the positive elements in the temp[] array. Then, we store all the negative elements from the original array into temp[]. Finally, we copy the contents of temp[] back into the original array, ensuring that the negative elements were placed at the end of the array.
C++
// C++ program to Move All -ve Element At End
// Without changing the relative ordering
// Using Naive Partitioning Algorithm
#include <iostream>
#include <vector>
using namespace std;
// Function to move all -ve element to end of array
// in same order.
void segregateElements(vector<int>& arr) {
int n = arr.size();
// Create an empty array to store result
vector<int> temp(n);
int idx = 0;
// First fill non-negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] >= 0)
temp[idx++] = arr[i];
}
// Now fill negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] < 0)
temp[idx++] = arr[i];
}
// copy the elements from temp to arr
arr = temp;
}
int main() {
vector<int> arr = {1 ,-1 ,-3 , -2, 7, 5, 11, 6};
segregateElements(arr);
for (int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// C program to Move All -ve Element At End
// Without changing the relative ordering
// Using Naive Partitioning Algorithm
#include <stdio.h>
#include <stdlib.h>
void segregateElements(int* arr, int n) {
// Create an empty array to store result
int* temp = (int*)malloc(n * sizeof(int));
int idx = 0;
// First fill non-negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] >= 0)
temp[idx++] = arr[i];
}
// Now fill negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] < 0)
temp[idx++] = arr[i];
}
// copy the elements from temp to arr
for (int i = 0; i < n; i++) {
arr[i] = temp[i];
}
free(temp);
}
int main() {
int arr[] = {1, -1, -3, -2, 7, 5, 11, 6};
int n = sizeof(arr) / sizeof(arr[0]);
segregateElements(arr, n);
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
// Java program to Move All -ve Element At End
// Without changing the relative ordering
// Using Naive Partitioning Algorithm
import java.util.Arrays;
class GfG {
// Function to move all -ve element to end of array
// in same order.
static void segregateElements(int[] arr) {
int n = arr.length;
// Create an empty array to store result
int[] temp = new int[n];
int idx = 0;
// First fill non-negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] >= 0)
temp[idx++] = arr[i];
}
// Now fill negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] < 0)
temp[idx++] = arr[i];
}
// copy the elements from temp to arr
System.arraycopy(temp, 0, arr, 0, n);
}
public static void main(String[] args) {
int[] arr = {1, -1, -3, -2, 7, 5, 11, 6};
segregateElements(arr);
for (int ele: arr)
System.out.print(ele + " ");
}
}
# Python program to Move All -ve Element At End
# Without changing the relative ordering
# Using Naive Partitioning Algorithm
def segregateElements(arr):
n = len(arr)
# Create an empty array to store result
temp = [0] * n
idx = 0
# First fill non-negative elements into the
# temporary array
for i in range(n):
if arr[i] >= 0:
temp[idx] = arr[i]
idx += 1
# Now fill negative elements into the
# temporary array
for i in range(n):
if arr[i] < 0:
temp[idx] = arr[i]
idx += 1
# copy the elements from temp to arr
for i in range(n):
arr[i] = temp[i]
if __name__ == "__main__":
arr = [1, -1, -3, -2, 7, 5, 11, 6]
segregateElements(arr)
print(" ".join(map(str, arr)))
// C# program to Move All -ve Element At End
// Without changing the relative ordering
// Using Naive Partitioning Algorithm
using System;
class GfG {
// Function to move all -ve element to end of array
// in same order.
static void segregateElements(int[] arr) {
int n = arr.Length;
// Create an empty array to store result
int[] temp = new int[n];
int idx = 0;
// First fill non-negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] >= 0)
temp[idx++] = arr[i];
}
// Now fill negative elements into the
// temporary array
for (int i = 0; i < n; i++) {
if (arr[i] < 0)
temp[idx++] = arr[i];
}
// copy the elements from temp to arr
Array.Copy(temp, arr, n);
}
static void Main() {
int[] arr = {1, -1, -3, -2, 7, 5, 11, 6};
segregateElements(arr);
Console.WriteLine(string.Join(" ", arr));
}
}
// JavaScript program to Move All -ve Element At End
// Without changing the relative ordering
// Using Naive Partitioning Algorithm
// Function to move all -ve element to end of array
// in same order.
function segregateElements(arr) {
let n = arr.length;
// Create an empty array to store result
let temp = new Array(n);
let idx = 0;
// First fill non-negative elements into the
// temporary array
for (let i = 0; i < n; i++) {
if (arr[i] >= 0)
temp[idx++] = arr[i];
}
// Now fill negative elements into the
// temporary array
for (let i = 0; i < n; i++) {
if (arr[i] < 0)
temp[idx++] = arr[i];
}
// copy the elements from temp to arr
for (let i = 0; i < n; i++) {
arr[i] = temp[i];
}
}
// Driver Code
let arr = [1, -1, -3, -2, 7, 5, 11, 6];
segregateElements(arr);
console.log(arr.join(" "));
Output1 7 5 11 6 -1 -3 -2
Time Complexity : O(n), since the array arr[] got traversed 3 times: first 2 traversal for filling the temporary array and 3rd traversal to copy elements back to original array.
Auxiliary space : O(n), space used by temporary array.
Using Two Stacks
The idea is to use two stacks to rearrange elements while preserving their relative order. The first stack stores negative elements, and the second stores non-negative elements. After processing the entire array, we refill it from the end: first with all negative elements from the negative stack and then with all non-negative elements from the positive stack. This ordering ensures that all negative elements are moved to the end while preserving the original sequence among positive values and among negative values.
C++
// C++ program to Move All -ve Element At End
// Without changing the relative ordering
// Using two Stacks
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
// Function to move all the negative elements
// to the end of the array in the same order
void segregateElements(vector<int> &arr) {
// Fill two stacks with positive and negative
// elements respectively
stack<int> neg, pos;
for (int i = 0; i < arr.size(); i++) {
if (arr[i] < 0)
neg.push(arr[i]);
else
pos.push(arr[i]);
}
// Place the elements back to original array arr[]
for (int i = arr.size()-1; i >= 0; i--) {
// first place the negative elements
if (!neg.empty()) {
arr[i] = neg.top();
neg.pop();
}
// If negative elements were all placed
// then place remaining elements
else {
arr[i] = pos.top();
pos.pop();
}
}
}
int main() {
vector<int> arr = {1, -1, -3, -2, 7, 5, 11, 6};
segregateElements(arr);
for (int ele: arr)
cout << ele << " ";
return 0;
}
// Java program to Move All -ve Element At End
// Without changing the relative ordering
// Using two Stacks
import java.util.Stack;
class GfG {
// Function to move all the negative elements
// to the end of the array in the same order
static void segregateElements(int[] arr) {
// Fill two stacks with positive and negative
// elements respectively
Stack<Integer> neg = new Stack<>();
Stack<Integer> pos = new Stack<>();
for (int value : arr) {
if (value < 0)
neg.push(value);
else
pos.push(value);
}
// Place the elements back to original array arr[]
for (int i = arr.length - 1; i >= 0; i--) {
// first place the negative elements
if (!neg.isEmpty()) {
arr[i] = neg.pop();
}
// If negative elements were all placed
// then place remaining elements
else {
arr[i] = pos.pop();
}
}
}
public static void main(String[] args) {
int[] arr = {1, -1, -3, -2, 7, 5, 11, 6};
segregateElements(arr);
for (int ele : arr)
System.out.print(ele + " ");
}
}
# Python program to Move All -ve Element At End
# Without changing the relative ordering
# Using two Stacks
# Function to move all the negative elements
# to the end of the array in the same order
def segregateElements(arr):
# Fill two stacks with positive and negative
# elements respectively
neg = []
pos = []
for value in arr:
if value < 0:
neg.append(value)
else:
pos.append(value)
# Place the elements back to original array arr
for i in range(len(arr) - 1, -1, -1):
# first place the negative elements
if neg:
arr[i] = neg.pop()
# If negative elements were all placed
# then place remaining elements
else:
arr[i] = pos.pop()
if __name__ == "__main__":
arr = [1, -1, -3, -2, 7, 5, 11, 6]
segregateElements(arr)
print(" ".join(map(str, arr)))
// C# program to Move All -ve Element At End
// Without changing the relative ordering
// Using two Stacks
using System;
using System.Collections.Generic;
class GfG {
// Function to move all the negative elements
// to the end of the array in the same order
static void segregateElements(int[] arr) {
// Fill two stacks with positive and negative
// elements respectively
Stack<int> neg = new Stack<int>();
Stack<int> pos = new Stack<int>();
foreach (int value in arr) {
if (value < 0)
neg.Push(value);
else
pos.Push(value);
}
// Place the elements back to original array arr[]
for (int i = arr.Length - 1; i >= 0; i--) {
// first place the negative elements
if (neg.Count > 0) {
arr[i] = neg.Pop();
}
// If negative elements were all placed
// then place remaining elements
else {
arr[i] = pos.Pop();
}
}
}
static void Main() {
int[] arr = {1, -1, -3, -2, 7, 5, 11, 6};
segregateElements(arr);
foreach (int ele in arr)
Console.Write(ele + " ");
}
}
// JavaScript program to Move All -ve Element At End
// Without changing the relative ordering
// Using two Stacks
// Function to move all the negative elements
// to the end of the array in the same order
function segregateElements(arr) {
// Fill two stacks with positive and negative
// elements respectively
let neg = [];
let pos = [];
for (let value of arr) {
if (value < 0)
neg.push(value);
else
pos.push(value);
}
// Place the elements back to original array arr[]
for (let i = arr.length - 1; i >= 0; i--) {
// first place the negative elements
if (neg.length > 0) {
arr[i] = neg.pop();
}
// If negative elements were all placed
// then place remaining elements
else {
arr[i] = pos.pop();
}
}
}
// Driver Code
let arr = [1, -1, -3, -2, 7, 5, 11, 6];
segregateElements(arr);
console.log(arr.join(" "));
Output1 7 5 11 6 -1 -3 -2
Time Complexity: O(n), since the array arr[] got traversed twice.
Auxiliary Space: O(n), used by the stacks.
Related Articles:
Rearrange positive and negative numbers with constant extra space
Alternate Rearrangement of Positives and Negatives
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