An array contains both positive and negative numbers in random order. Rearrange the array elements so that all negative numbers appear before all positive numbers.
Examples :
Input: -12, 11, -13, -5, 6, -7, 5, -3, -6 Output: -12 -13 -5 -7 -3 -6 11 6 5
Note: Order of elements is not important here.
Approach 1:
The idea is to simply apply the partition process of quicksort.
C++
// A C++ program to put all negative // numbers before positive numbers #include <bits/stdc++.h> using namespace std; void rearrange( int arr[], int n) { int j = 0; for ( int i = 0; i < n; i++) { if (arr[i] < 0) { if (i != j) swap(arr[i], arr[j]); j++; } } } // A utility function to print an array void printArray( int arr[], int n) { for ( int i = 0; i < n; i++) printf ( "%d " , arr[i]); } // Driver code int 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
// Java program to put all negative // numbers before positive numbers import java.io.*; class GFG { static void rearrange( int arr[], int n) { int j = 0 , temp; for ( int i = 0 ; i < n; i++) { if (arr[i] < 0 ) { if (i != j) { temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } j++; } } } // 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 code public static void main(String args[]) { 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 Nikita Tiwari. |
Python3
# A Python 3 program to put # all negative numbers before # positive numbers def rearrange(arr, n ) : # Please refer partition() in # below post # https://www.geeksforgeeks.org / quick-sort / j = 0 j = 0 for i in range ( 0 , n) : if (arr[i] < 0 ) : temp = arr[i] arr[i] = arr[j] arr[j] = temp j = j + 1 print (arr) # Driver code arr = [ - 1 , 2 , - 3 , 4 , 5 , 6 , - 7 , 8 , 9 ] n = len (arr) rearrange(arr, n) # This code is contributed by Nikita Tiwari. |
C#
// C# program to put all negative // numbers before positive numbers using System; class GFG { static void rearrange( int [] arr, int n) { int j = 0, temp; for ( int i = 0; i < n; i++) { if (arr[i] < 0) { temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; j++; } } } // 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 code 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 nitin mittal. |
PHP
<?php // A PHP program to put all negative // numbers before positive numbers function rearrange(& $arr , $n ) { $j = 0; for ( $i = 0; $i < $n ; $i ++) { if ( $arr [ $i ] < 0) { if ( $i != $j ) { $temp = $arr [ $i ]; $arr [ $i ] = $arr [ $j ]; $arr [ $j ] = $temp ; } $j ++; } } } // 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 = sizeof( $arr ); rearrange( $arr , $n ); printArray( $arr , $n ); // This code is contributed by ChitraNayal ?> |
-1 -3 -7 4 5 6 2 8 9
Time complexity: O(N)
Auxiliary Space: O(1)
Two Pointer Approach: The idea is to solve this problem with constant space and linear time is by using a two-pointer or two-variable approach where we simply take two variables like left and right which hold the 0 and N-1 indexes. Just need to check that :
- Check If the left and right pointer elements are negative then simply increment the left pointer.
- Otherwise, if the left element is positive and the right element is negative then simply swap the elements, and Simultaneously increment or decrement the left and right pointers.
- Else if the left element is positive and the right element is also positive then simply decrement the right pointer.
- Repeat the above 3 steps until the left pointer ≤ right pointer.
Below is the implementation of the above approach:
C++
// C++ program of the above // approach #include <iostream> using namespace std; // Function to shift all the // negative elements on left side void shiftall( int arr[], int left, int right) { // Loop to iterate over the // array from left to the right while (left<=right) { // Condition to check if the left // and the right elements are // negative if (arr[left] < 0 && arr[right] < 0) left+=1; // Condition to check if the left // pointer element is positive and // the right pointer element is negative else if (arr[left]>0 && arr[right]<0) { int temp=arr[left]; arr[left]=arr[right]; arr[right]=temp; left+=1; right-=1; } // Condition to check if both the // elements are positive else if (arr[left]>0 && arr[right] >0) right-=1; else { left += 1; right -= 1; } } } // Function to print the array void display( int arr[], int right){ // Loop to iterate over the element // of the given array for ( int i=0;i<=right;++i){ cout<<arr[i]<< " " ; } cout<<endl; } // Driver Code int main() { int arr[] = {-12, 11, -13, -5, 6, -7, 5, -3, 11}; int arr_size = sizeof (arr) / sizeof (arr[0]); // Function Call shiftall(arr,0,arr_size-1); display(arr,arr_size-1); return 0; } //added by Dhruv Goyal |
Java
// Java program of the above // approach import java.io.*; class GFG{ // Function to shift all the // negative elements on left side static void shiftall( int [] arr, int left, int right) { // Loop to iterate over the // array from left to the right while (left <= right) { // Condition to check if the left // and the right elements are // negative if (arr[left] < 0 && arr[right] < 0 ) left++; // Condition to check if the left // pointer element is positive and // the right pointer element is negative else if (arr[left] > 0 && arr[right] < 0 ) { int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } // Condition to check if both the // elements are positive else if (arr[left] > 0 && arr[right] > 0 ) right--; else { left++; right--; } } } // Function to print the array static void display( int [] arr, int right) { // Loop to iterate over the element // of the given array for ( int i = 0 ; i <= right; ++i) System.out.print(arr[i] + " " ); System.out.println(); } // Drive code public static void main(String[] args) { int [] arr = { - 12 , 11 , - 13 , - 5 , 6 , - 7 , 5 , - 3 , 11 }; int arr_size = arr.length; // Function Call shiftall(arr, 0 , arr_size - 1 ); display(arr, arr_size - 1 ); } } // This code is contributed by dhruvgoyal267 |
Python3
# Python3 program of the # above approach # Function to shift all the # the negative elements to # the left of the array def shiftall(arr,left,right): # Loop to iterate while the # left pointer is less than # the right pointer while left< = right: # Condition to check if the left # and right pointer negative if arr[left] < 0 and arr[right] < 0 : left + = 1 # Condition to check if the left # pointer element is positive and # the right pointer element is # negative elif arr[left]> 0 and arr[right]< 0 : arr[left], arr[right] = \ arr[right],arr[left] left + = 1 right - = 1 # Condition to check if the left # pointer is positive and right # pointer as well elif arr[left]> 0 and arr[right]> 0 : right - = 1 else : left + = 1 right - = 1 # Function to print the array def display(arr): for i in range ( len (arr)): print (arr[i], end = " " ) print () # Driver Code if __name__ = = "__main__" : arr = [ - 12 , 11 , - 13 , - 5 , \ 6 , - 7 , 5 , - 3 , 11 ] n = len (arr) shiftall(arr, 0 ,n - 1 ) display(arr) # Sumit Singh |
C#
// C# program of the above // approach using System.IO; using System; class GFG { // Function to shift all the // negative elements on left side static void shiftall( int [] arr, int left, int right) { // Loop to iterate over the // array from left to the right while (left <= right) { // Condition to check if the left // and the right elements are // negative if (arr[left] < 0 && arr[right] < 0) left++; // Condition to check if the left // pointer element is positive and // the right pointer element is negative else if (arr[left] > 0 && arr[right] < 0) { int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } // Condition to check if both the // elements are positive else if (arr[left] > 0 && arr[right] > 0) right--; else { left++; right--; } } } // Function to print the array static void display( int [] arr, int right) { // Loop to iterate over the element // of the given array for ( int i = 0; i <= right; ++i) { Console.Write(arr[i] + " " ); } Console.WriteLine(); } // Drive code static void Main() { int [] arr = {-12, 11, -13, -5, 6, -7, 5, -3, 11}; int arr_size = arr.Length; shiftall(arr, 0, arr_size - 1); display(arr, arr_size - 1); } } // This code is contributed by avanitrachhadiya2155 |
-12 -3 -13 -5 -7 6 5 11 11
This is an in-place rearranging algorithm for arranging the positive and negative numbers where the order of elements is not maintained.
Time Complexity: O(N)
Auxiliary Space: O(1)
The problem becomes difficult if we need to maintain the order of elements. Please refer to Rearrange positive and negative numbers with constant extra space for details.
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