An array contains both positive and negative numbers in random order. Rearrange the array elements so that positive and negative numbers are placed alternatively. A number of positive and negative numbers need not be equal. If there are more positive numbers they appear at the end of the array. If there are more negative numbers, they too appear at the end of the array.
Example:
Input: [-1, 2, -3, 4, 5, 6, -7, 8, 9]
Output:[9, -7, 8, -3, 5, -1, 2, 4, 6]Input: [-1, 3, -2, -4, 7, -5]
Output:[7, -2, 3, -5, -1, -4]
Note: The partition process in this approach changes the relative order of elements. I.e. the order of the appearance of elements is not maintained with this approach. See this for maintaining the order of appearance of elements in this problem.
Approach:
The solution is to first separate positive and negative numbers using the partition process of QuickSort. In the partition process, consider 0 as the value of the pivot element so that all negative numbers are placed before positive numbers. Once negative and positive numbers are separated, we start from the first negative number and first positive number and swap every alternate negative number with the next positive number.
Below is the implementation of above idea:
C++
// A C++ program to put positive// numbers at even indexes (0, 2, 4,..) // and negative numbers at odd // indexes (1, 3, 5, ..)#include <iostream>using namespace std;class GFG{ public: void rearrange(int [],int); void swap(int *,int *); void printArray(int [],int);};// The main function that rearranges // elements of given array. It puts// positive elements at even indexes // (0, 2, ..) and negative numbers // at odd indexes (1, 3, ..).void GFG :: rearrange(int arr[], int n){ // The following few lines are // similar to partition process // of QuickSort. The idea is to // consider 0 as pivot and // divide the array around it. int i = -1; for (int j = 0; j < n; j++) { if (arr[j] < 0) { i++; swap(&arr[i], &arr[j]); } } // Now all positive numbers are at // end and negative numbers at the // beginning of array. Initialize // indexes for starting point of // positive and negative numbers // to be swapped int pos = i + 1, neg = 0; // Increment the negative index by // 2 and positive index by 1, // i.e., swap every alternate negative // number with next positive number while (pos < n && neg < pos && arr[neg] < 0) { swap(&arr[neg], &arr[pos]); pos++; neg += 2; }}// A utility function // to swap two elementsvoid GFG :: swap(int *a, int *b){ int temp = *a; *a = *b; *b = temp;}// A utility function to print an arrayvoid GFG :: printArray(int arr[], int n){ for (int i = 0; i < n; i++) cout << arr[i] << " ";}// Driver Codeint main() { int arr[] = {-1, 2, -3, 4, 5, 6, -7, 8, 9}; int n = sizeof(arr) / sizeof(arr[0]); GFG test; test.rearrange(arr, n); test.printArray(arr, n); return 0;}// This code is contributed // by vt_Yogesh Shukla 1 |
C
// A C++ program to put positive numbers at even indexes (0, // 2, 4,..) and negative numbers at odd indexes (1, 3, 5, ..)#include <stdio.h>// prototype for swapvoid swap(int *a, int *b);// The main function that rearranges elements of given array. // It puts positive elements at even indexes (0, 2, ..) and // negative numbers at odd indexes (1, 3, ..).void rearrange(int arr[], int n){ // The following few lines are similar to partition process // of QuickSort. The idea is to consider 0 as pivot and // divide the array around it. int i = -1; for (int j = 0; j < n; j++) { if (arr[j] < 0) { i++; swap(&arr[i], &arr[j]); } } // Now all positive numbers are at end and negative numbers // at the beginning of array. Initialize indexes for starting // point of positive and negative numbers to be swapped int pos = i+1, neg = 0; // Increment the negative index by 2 and positive index by 1, // i.e., swap every alternate negative number with next // positive number while (pos < n && neg < pos && arr[neg] < 0) { swap(&arr[neg], &arr[pos]); pos++; neg += 2; }}// A utility function to swap two elementsvoid swap(int *a, int *b){ int temp = *a; *a = *b; *b = temp;}// A utility function to print an arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) printf("%4d ", arr[i]);}// Driver program to test above functionsint main(){ int arr[] = {-1, 2, -3, 4, 5, 6, -7, 8, 9}; int n = sizeof(arr)/sizeof(arr[0]); rearrange(arr, n); printArray(arr, n); return 0;} |
Java
// A JAVA program to put positive numbers at even indexes// (0, 2, 4,..) and negative numbers at odd indexes (1, 3,// 5, ..)import java.io.*;class Alternate { // The main function that rearranges elements of given // array. It puts positive elements at even indexes (0, // 2, ..) and negative numbers at odd indexes (1, 3, ..). static void rearrange(int arr[], int n) { // The following few lines are similar to partition // process of QuickSort. The idea is to consider 0 // as pivot and divide the array around it. int i = -1, temp = 0; for (int j = 0; j < n; j++) { if (arr[j] < 0) { i++; temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } // Now all positive numbers are at end and negative numbers at // the beginning of array. Initialize indexes for starting point // of positive and negative numbers to be swapped int pos = i+1, neg = 0; // Increment the negative index by 2 and positive index by 1, i.e., // swap every alternate negative number with next positive number while (pos < n && neg < pos && arr[neg] < 0) { temp = arr[neg]; arr[neg] = arr[pos]; arr[pos] = temp; pos++; neg += 2; } } // A utility function to print an array static void printArray(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } /*Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {-1, 2, -3, 4, 5, 6, -7, 8, 9}; int n = arr.length; rearrange(arr,n); System.out.println("Array after rearranging: "); printArray(arr,n); }}/*This code is contributed by Devesh Agrawal*/ |
Python3
# Python program to put positive numbers at even indexes (0, // 2, 4,..) and# negative numbers at odd indexes (1, 3, 5, ..)# The main function that rearranges elements of given array. # It puts positive elements at even indexes (0, 2, ..) and # negative numbers at odd indexes (1, 3, ..).def rearrange(arr, n): # The following few lines are similar to partition process # of QuickSort. The idea is to consider 0 as pivot and # divide the array around it. i = -1 for j in range(n): if (arr[j] < 0): i += 1 # swapping of arr arr[i], arr[j] = arr[j], arr[i] # Now all positive numbers are at end and negative numbers # at the beginning of array. Initialize indexes for starting # point of positive and negative numbers to be swapped pos, neg = i+1, 0 # Increment the negative index by 2 and positive index by 1, # i.e., swap every alternate negative number with next # positive number while (pos < n and neg < pos and arr[neg] < 0): # swapping of arr arr[neg], arr[pos] = arr[pos], arr[neg] pos += 1 neg += 2# A utility function to print an arraydef printArray(arr, n): for i in range(n): print (arr[i],end=" ") # Driver program to test above functionsarr = [-1, 2, -3, 4, 5, 6, -7, 8, 9]n = len(arr)rearrange(arr, n)printArray(arr, n)# Contributed by Afzal |
C#
// A C# program to put positive numbers// at even indexes (0, 2, 4, ..) and // negative numbers at odd indexes (1, 3, 5, ..)using System;class Alternate { // The main function that rearranges elements // of given array. It puts positive elements // at even indexes (0, 2, ..) and negative // numbers at odd indexes (1, 3, ..). static void rearrange(int[] arr, int n) { // The following few lines are similar to partition // process of QuickSort. The idea is to consider 0 // as pivot and divide the array around it. int i = -1, temp = 0; for (int j = 0; j < n; j++) { if (arr[j] < 0) { i++; temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } // Now all positive numbers are at end // and negative numbers at the beginning of // array. Initialize indexes for starting point // of positive and negative numbers to be swapped int pos = i + 1, neg = 0; // Increment the negative index by 2 and // positive index by 1, i.e., swap every // alternate negative number with next positive number while (pos < n && neg < pos && arr[neg] < 0) { temp = arr[neg]; arr[neg] = arr[pos]; arr[pos] = temp; pos++; neg += 2; } } // A utility function to print an array static void printArray(int[] arr, int n) { for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } /*Driver function to check for above functions*/ public static void Main() { int[] arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 }; int n = arr.Length; rearrange(arr, n); printArray(arr, n); }}// This code is contributed by vt_m. |
Javascript
<script>// Javascript program to put positive// numbers at even indexes// (0, 2, 4,..) and negative // numbers at odd indexes (1, 3,// 5, ..) // The main function that // rearranges elements of given // array. It puts positive elements // at even indexes (0, // 2, ..) and negative numbers // at odd indexes (1, 3, ..). function rearrange(arr,n) { // The following few lines are similar to partition // process of QuickSort. The idea is to consider 0 // as pivot and divide the array around it. let i = -1, temp = 0; for (let j = 0; j < n; j++) { if (arr[j] < 0) { i++; temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } // Now all positive numbers are // at end and negative numbers at // the beginning of array. // Initialize indexes for starting point // of positive and negative numbers // to be swapped let pos = i+1, neg = 0; // Increment the negative index // by 2 and positive index by 1, i.e., // swap every alternate negative number // with next positive number while (pos < n && neg < pos && arr[neg] < 0) { temp = arr[neg]; arr[neg] = arr[pos]; arr[pos] = temp; pos++; neg += 2; } } // A utility function to print an array function printArray(arr,n) { for (let i = 0; i < n; i++) document.write(arr[i] + " "); } /*Driver function to check for above functions*/ let arr = [-1, 2, -3, 4, 5, 6, -7, 8, 9]; let n = arr.length; rearrange(arr,n); printArray(arr,n); // This code is contributed by sravan kumar</script> |
PHP
<?php// A PHP program to put positive numbers // at even indexes (0, 2, 4,..) and negative // numbers at odd indexes (1, 3, 5, ..) // The main function that rearranges elements // of given array. It puts positive elements // at even indexes (0, 2, ..) and negative// numbers at odd indexes (1, 3, ..). function rearrange(&$arr, $n) { // The following few lines are similar // to partition process of QuickSort. // The idea is to consider 0 as pivot // and divide the array around it. $i = -1; for ($j = 0; $j < $n; $j++) { if ($arr[$j] < 0) { $i++; swap($arr[$i], $arr[$j]); } } // Now all positive numbers are at end and // negative numbers at the beginning of array. // Initialize indexes for starting point of // positive and negative numbers to be swapped $pos = $i + 1; $neg = 0; // Increment the negative index by 2 and // positive index by 1, i.e., swap every // alternate negative number with next // positive number while ($pos < $n && $neg < $pos && $arr[$neg] < 0) { swap($arr[$neg], $arr[$pos]); $pos++; $neg += 2; } } // A utility function to swap two elements function swap(&$a, &$b) { $temp = $a; $a = $b; $b = $temp; } // A utility function to print an array function printArray(&$arr, $n) { for ($i = 0; $i < $n; $i++) echo " " . $arr[$i] . " "; } // Driver Code$arr = array(-1, 2, -3, 4, 5, 6, -7, 8, 9); $n = count($arr); rearrange($arr, $n); printArray($arr, $n); // This code is contributed// by rathbhupendra?> |
Output:
4 -3 5 -1 6 -7 2 8 9
Time Complexity: O(n) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(1), as we are not using any extra space.
Related Articles:
Rearrange positive and negative numbers with constant extra space
Move all negative elements to end in order with extra space allowed
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