Rearrange array in alternating positive & negative items with O(1) extra space | Set 2
Given an array of positive and negative numbers, arrange them in an alternate fashion such that every positive number is followed by negative and vice-versa. Order of elements in output doesn’t matter. Extra positive or negative elements should be moved to end.
Examples:
Input : arr[] = {-2, 3, 4, -1} Output : arr[] = {-2, 3, -1, 4} OR {-1, 3, -2, 4} OR .. Input : arr[] = {-2, 3, 1} Output : arr[] = {-2, 3, 1} OR {-2, 1, 3} Input : arr[] = {-5, 3, 4, 5, -6, -2, 8, 9, -1, -4} Output : arr[] = {-5, 3, -2, 5, -6, 4, -4, 9, -1, 8} OR ..
Approach 1:
- First, sort the array in non-increasing order. Then we will count the number of positive and negative integers.
- Then swap the one negative and one positive number in the odd positions till we reach our condition.
- This will rearrange the array elements because we are sorting the array and accessing the element from left to right according to our need.
Below is the implementation of the above approach:
C++
// Below is the implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function which works in the condition when number of // negative numbers are lesser or equal than positive // numbers void fill1( int a[], int neg, int pos) { if (neg % 2 == 1) { for ( int i = 1; i < neg; i += 2) { int c = a[i]; int d = a[i + neg]; int temp = c; a[i] = d; a[i + neg] = temp; } } else { for ( int i = 1; i <= neg; i += 2) { int c = a[i]; int d = a[i + neg - 1]; int temp = c; a[i] = d; a[i + neg - 1] = temp; } } } // Function which works in the condition when number of // negative numbers are greater than positive numbers void fill2( int a[], int neg, int pos) { if (pos % 2 == 1) { for ( int i = 1; i < pos; i += 2) { int c = a[i]; int d = a[i + pos]; int temp = c; a[i] = d; a[i + pos] = temp; } } else { for ( int i = 1; i <= pos; i += 2) { int c = a[i]; int d = a[i + pos - 1]; int temp = c; a[i] = d; a[i + pos - 1] = temp; } } } // Reverse the array void reverse( int a[], int n) { int i, k, t; for (i = 0; i < n / 2; i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } } // Print the array void print( int a[], int n) { for ( int i = 0; i < n; i++) cout << a[i] << " " ; cout << endl; } // Driver code int main() { int arr[] = { 2, 3, -4, -1, 6, -9 }; int n = sizeof (arr) / sizeof (arr[0]); cout << "Given array is " ; print(arr, n); int neg = 0, pos = 0; for ( int i = 0; i < n; i++) { if (arr[i] < 0) neg++; else pos++; } // Sort the array sort(arr, arr + n); if (neg <= pos) fill1(arr, neg, pos); else { // Reverse the array in this condition reverse(arr, n); fill2(arr, neg, pos); } cout << "Rearranged array is " ; print(arr, n); } // This code is contributed by Potta Lokesh |
C
// Below is the implementation of the above approach #include <stdio.h> #include <stdlib.h> // Compare function for qsort int cmpfunc( const void * a, const void * b) { return (*( int *)a - *( int *)b); } // Function which works in the condition when number of // negative numbers are lesser or equal than positive // numbers void fill1( int a[], int neg, int pos) { if (neg % 2 == 1) { for ( int i = 1; i < neg; i += 2) { int c = a[i]; int d = a[i + neg]; int temp = c; a[i] = d; a[i + neg] = temp; } } else { for ( int i = 1; i <= neg; i += 2) { int c = a[i]; int d = a[i + neg - 1]; int temp = c; a[i] = d; a[i + neg - 1] = temp; } } } // Function which works in the condition when number of // negative numbers are greater than positive numbers void fill2( int a[], int neg, int pos) { if (pos % 2 == 1) { for ( int i = 1; i < pos; i += 2) { int c = a[i]; int d = a[i + pos]; int temp = c; a[i] = d; a[i + pos] = temp; } } else { for ( int i = 1; i <= pos; i += 2) { int c = a[i]; int d = a[i + pos - 1]; int temp = c; a[i] = d; a[i + pos - 1] = temp; } } } // Reverse the array void reverse( int a[], int n) { int i, k, t; for (i = 0; i < n / 2; i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } } // Print the array void print( int a[], int n) { for ( int i = 0; i < n; i++) printf ( "%d " , a[i]); printf ( "\n" ); } // Driver code int main() { int arr[] = { 2, 3, -4, -1, 6, -9 }; int n = sizeof (arr) / sizeof (arr[0]); printf ( "Given array is " ); print(arr, n); int neg = 0, pos = 0; for ( int i = 0; i < n; i++) { if (arr[i] < 0) neg++; else pos++; } // Sort the array qsort (arr, n, sizeof ( int ), cmpfunc); if (neg <= pos) fill1(arr, neg, pos); else { // Reverse the array in this condition reverse(arr, n); fill2(arr, neg, pos); } printf ( "Rearranged array is " ); print(arr, n); } // This code is contributed by Sania Kumari Gupta |
Java
// Below is the implementation of the above approach import java.io.*; import java.lang.*; import java.util.*; public class Main { // function which works in the condition when number of // negative numbers are lesser or equal than positive // numbers static void fill1( int a[], int neg, int pos) { if (neg % 2 == 1 ) { for ( int i = 1 ; i < neg; i += 2 ) { int c = a[i]; int d = a[i + neg]; int temp = c; a[i] = d; a[i + neg] = temp; } } else { for ( int i = 1 ; i <= neg; i += 2 ) { int c = a[i]; int d = a[i + neg - 1 ]; int temp = c; a[i] = d; a[i + neg - 1 ] = temp; } } } // Function which works in the condition when number of // negative numbers are greater than positive numbers static void fill2( int a[], int neg, int pos) { if (pos % 2 == 1 ) { for ( int i = 1 ; i < pos; i += 2 ) { int c = a[i]; int d = a[i + pos]; int temp = c; a[i] = d; a[i + pos] = temp; } } else { for ( int i = 1 ; i <= pos; i += 2 ) { int c = a[i]; int d = a[i + pos - 1 ]; int temp = c; a[i] = d; a[i + pos - 1 ] = temp; } } } // Reverse the array static void reverse( int a[], int n) { int i, k, t; for (i = 0 ; i < n / 2 ; i++) { t = a[i]; a[i] = a[n - i - 1 ]; a[n - i - 1 ] = t; } } // Print the array static void print( int a[], int n) { for ( int i = 0 ; i < n; i++) System.out.print(a[i] + " " ); System.out.println(); } // Driver Code public static void main(String[] args) throws java.lang.Exception { // Given array int [] arr = { 2 , 3 , - 4 , - 1 , 6 , - 9 }; int n = arr.length; System.out.println( "Given array is " ); print(arr, n); int neg = 0 , pos = 0 ; for ( int i = 0 ; i < n; i++) { if (arr[i] < 0 ) neg++; else pos++; } // Sort the array Arrays.sort(arr); if (neg <= pos) { fill1(arr, neg, pos); } else { // reverse the array in this condition reverse(arr, n); fill2(arr, neg, pos); } System.out.println( "Rearranged array is " ); print(arr, n); } } |
Python3
# Python3 program for the above approach # Function which works in the condition # when number of negative numbers are # lesser or equal than positive numbers def fill1(a, neg, pos) : if (neg % 2 = = 1 ) : for i in range ( 1 , neg, 2 ): c = a[i] d = a[i + neg] temp = c a[i] = d a[i + neg] = temp else : for i in range ( 1 , neg + 1 , 2 ): c = a[i] d = a[i + neg - 1 ] temp = c a[i] = d a[i + neg - 1 ] = temp # Function which works in the condition # when number of negative numbers are # greater than positive numbers def fill2(a, neg, pos): if (pos % 2 = = 1 ) : for i in range ( 1 , pos, 2 ): c = a[i] d = a[i + pos] temp = c a[i] = d a[i + pos] = temp else : for i in range ( 1 , pos + 1 , 2 ): c = a[i] d = a[i + pos - 1 ] temp = c a[i] = d a[i + pos - 1 ] = temp # Reverse the array def reverse(a, n) : for i in range (n / 2 ): t = a[i] a[i] = a[n - i - 1 ] a[n - i - 1 ] = t # Print the array def printt(a, n): for i in range (n): print (a[i], end = " " ) print () # Driver code if __name__ = = "__main__" : arr = [ 2 , 3 , - 4 , - 1 , 6 , - 9 ] n = len (arr) print ( "Given array is " ) printt(arr, n) neg = 0 pos = 0 for i in range ( 0 , n): if (arr[i] < 0 ): neg + = 1 else : pos + = 1 # Sort the array arr.sort() if (neg < = pos) : fill1(arr, neg, pos) else : # Reverse the array in this condition reverse(arr, n) fill2(arr, neg, pos) print ( "Rearranged array is " ) printt(arr, n) # This code is contributed by sanjoy_62. |
C#
// C# program for the above approach using System; using System.Collections.Generic; using System.Linq; public class GFG { // function which works in the condition when number of // negative numbers are lesser or equal than positive // numbers static void fill1( int [] a, int neg, int pos) { if (neg % 2 == 1) { for ( int i = 1; i < neg; i += 2) { int c = a[i]; int d = a[i + neg]; int temp = c; a[i] = d; a[i + neg] = temp; } } else { for ( int i = 1; i <= neg; i += 2) { int c = a[i]; int d = a[i + neg - 1]; int temp = c; a[i] = d; a[i + neg - 1] = temp; } } } // Function which works in the condition when number of // negative numbers are greater than positive numbers static void fill2( int [] a, int neg, int pos) { if (pos % 2 == 1) { for ( int i = 1; i < pos; i += 2) { int c = a[i]; int d = a[i + pos]; int temp = c; a[i] = d; a[i + pos] = temp; } } else { for ( int i = 1; i <= pos; i += 2) { int c = a[i]; int d = a[i + pos - 1]; int temp = c; a[i] = d; a[i + pos - 1] = temp; } } } // Reverse the array static void reverse( int [] a, int n) { int i, k, t; for (i = 0; i < n / 2; i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } } // Print the array static void print( int [] a, int n) { for ( int i = 0; i < n; i++) Console.Write(a[i] + " " ); Console.WriteLine(); } // Driver Code public static void Main ( string [] args) { // Given array int [] arr = { 2, 3, -4, -1, 6, -9 }; int n = arr.Length; Console.WriteLine( "Given array is " ); print(arr, n); int neg = 0, pos = 0; for ( int i = 0; i < n; i++) { if (arr[i] < 0) neg++; else pos++; } // Sort the array Array.Sort(arr); if (neg <= pos) { fill1(arr, neg, pos); } else { // reverse the array in this condition reverse(arr, n); fill2(arr, neg, pos); } Console.WriteLine( "Rearranged array is " ); print(arr, n); } } // This code is contributed by splevel62. |
Javascript
<script> // Below is the implementation of the above approach // Function which works in the condition // when number of negative numbers are // lesser or equal than positive numbers function fill1(a, neg, pos) { if (neg % 2 == 1) { for (let i = 1; i < neg; i += 2) { let c = a[i]; let d = a[i + neg]; let temp = c; a[i] = d; a[i + neg] = temp; } } else { for (let i = 1; i <= neg; i += 2) { let c = a[i]; let d = a[i + neg - 1]; let temp = c; a[i] = d; a[i + neg - 1] = temp; } } } // Function which works in the condition // when number of negative numbers are // greater than positive numbers function fill2(a, neg, pos) { if (pos % 2 == 1) { for (let i = 1; i < pos; i += 2) { let c = a[i]; let d = a[i + pos]; let temp = c; a[i] = d; a[i + pos] = temp; } } else { for (let i = 1; i <= pos; i += 2) { let c = a[i]; let d = a[i + pos - 1]; let temp = c; a[i] = d; a[i + pos - 1] = temp; } } } // Reverse the array function reverse(a, n) { let i, k, t; for (i = 0; i < parseInt(n / 2); i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } } // Print the array function print(a, n) { for (let i = 0; i < n; i++) document.write(a[i] + " " ); document.write( '<br>' ); } // Driver code var arr = [ 2, 3, -4, -1, 6, -9 ]; let n = arr.length; document.write( "Given array is " ); print(arr, n); let neg = 0, pos = 0; for (let i = 0; i < n; i++) { if (arr[i] < 0) neg++; else pos++; } // Sort the array arr.sort( function (a, b){ return a - b;}); if (neg <= pos) { fill1(arr, neg, pos); } else { // Reverse the array in this condition reverse(arr, n); fill2(arr, neg, pos); } document.write( "Rearranged array is " ); print(arr, n); // This code is contributed by Potta Lokesh </script> |
Given array is 2 3 -4 -1 6 -9 Rearranged array is -9 3 -1 2 -4 6
Time Complexity: O(N*logN)
Space Complexity: O(1)
Efficient Approach: We have already discussed a O(n2) solution that maintains the order of appearance in the array here. If we are allowed to change order of appearance, we can solve this problem in O(n) time and O(1) space.
The idea is to process the array and shift all negative values to the end in O(n) time. After all negative values are shifted to the end, we can easily rearrange array in alternating positive & negative items. We basically swap next positive element at even position from next negative element in this step.
Following is the implementation of above idea.
C++
// C++ program to rearrange // array in alternating // positive & negative items // with O(1) extra space #include <bits/stdc++.h> using namespace std; // Function to rearrange positive and negative // integers in alternate fashion. The below // solution doesn't maintain original order of // elements void rearrange( int arr[], int n) { int i = 0, j = n-1; // shift all negative values to the end while (i < j) { while (i <= n - 1 and arr[i] > 0) i += 1; while (j >= 0 and arr[j] < 0) j -= 1; if (i < j ) swap(arr[i], arr[j]); } // i has index of leftmost // negative element if (i == 0 || i == n) return ; // start with first positive // element at index 0 // Rearrange array in alternating // positive & // negative items int k = 0; while (k < n && i < n ) { // swap next positive // element at even position // from next negative element. swap(arr[k], arr[i]); i = i + 1; k = k + 2; } } // Utility function to print an array void printArray( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; cout << endl; } // Driver code int main() { int arr[] = { 2, 3, -4, -1, 6, -9 }; int n = sizeof (arr) / sizeof (arr[0]); cout << "Given array is \n" ; printArray(arr, n); rearrange(arr, n); cout << "Rearranged array is \n" ; printArray(arr, n); return 0; } |
Java
// Java program to rearrange // array in alternating // positive & negative // items with O(1) extra space class GFG { // Function to rearrange positive and negative // integers in alternate fashion. The below // solution doesn't maintain original order of // elements static void rearrange( int arr[], int n) { int i = 0 , j = n - 1 ; // shift all negative values to the end while (i < j) { while (i <= n - 1 && arr[i] > 0 ) i += 1 ; while (j >= 0 && arr[j] < 0 ) j -= 1 ; if (i < j) swap(arr, i, j); } // i has index of leftmost negative element if (i == 0 || i == n) return ; // start with first positive // element at index 0 // Rearrange array in alternating positive & // negative items int k = 0 ; while (k < n && i < n) { // swap next positive element // at even position // from next negative element. swap(arr, k, i); i = i + 1 ; k = k + 2 ; } } // 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] + " " ); System.out.println( "" ); } static void swap( int arr[], int index1, int index2) { int c = arr[index1]; arr[index1] = arr[index2]; arr[index2] = c; } // Driver code public static void main(String[] args) { int arr[] = { 2 , 3 , - 4 , - 1 , 6 , - 9 }; int n = arr.length; System.out.println( "Given array is " ); printArray(arr, n); rearrange(arr, n); System.out.println( "Rearranged array is " ); printArray(arr, n); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to rearrange array # in alternating positive & negative # items with O(1) extra space # Function to rearrange positive and # negative integers in alternate fashion. # The below solution does not maintain # original order of elements def rearrange(arr, n): i = 0 j = n - 1 # shift all negative values # to the end while (i < j): while (i < = n - 1 and arr[i] > 0 ): i + = 1 while (j > = 0 and arr[j] < 0 ): j - = 1 if (i < j): temp = arr[i] arr[i] = arr[j] arr[j] = temp # i has index of leftmost # negative element if (i = = 0 or i = = n): return 0 # start with first positive element # at index 0 # Rearrange array in alternating # positive & negative items k = 0 while (k < n and i < n): # swap next positive element at even # position from next negative element. temp = arr[k] arr[k] = arr[i] arr[i] = temp i = i + 1 k = k + 2 # Utility function to print an array def printArray(arr, n): for i in range (n): print (arr[i], end = " " ) print ( "\n" ) # Driver code arr = [ 2 , 3 ] n = len (arr) print ( "Given array is" ) printArray(arr, n) rearrange(arr, n) print ( "Rearranged array is" ) printArray(arr, n) # This code is contributed # Princi Singh |
C#
// C# program to rearrange array // in alternating positive & negative // items with O(1) extra space using System; class GFG { // Function to rearrange positive and // negative integers in alternate fashion. // The below solution doesn't maintain // original order of elements static void rearrange( int [] arr, int n) { int i = 0, j = n - 1; // shift all negative values // to the end while (i < j) { while (i <= n - 1 && arr[i] > 0) i += 1; while (j >= 0 && arr[j] < 0) j -= 1; if (i < j) swap(arr, i, j); } // i has index of leftmost // negative element if (i == 0 || i == n) return ; // start with first positive // element at index 0 // Rearrange array in alternating // positive & negative items int k = 0; while (k < n && i < n) { // swap next positive element // at even position from next // negative element. swap(arr, k, i); i = i + 1; k = k + 2; } } // Utility function to print an array static void printArray( int [] arr, int n) { for ( int i = 0; i < n; i++) Console.Write(arr[i] + " " ); Console.WriteLine( "" ); } static void swap( int [] arr, int index1, int index2) { int c = arr[index1]; arr[index1] = arr[index2]; arr[index2] = c; } // Driver code public static void Main() { int [] arr = { 2, 3, -4, -1, 6, -9 }; int n = arr.Length; Console.WriteLine( "Given array is " ); printArray(arr, n); rearrange(arr, n); Console.WriteLine( "Rearranged array is " ); printArray(arr, n); } } // This code is contributed // by 29AjayKumar |
PHP
<?php // PHP program to rearrange array // in alternating positive & negative // items with O(1) extra space // Function to rearrange positive and // negative integers in alternate fashion. // The below solution doesn't maintain // original order of elements function rearrange(& $arr , $n ) { $i = 0; $j = $n - 1; // shift all negative values // to the end while ( $i < $j ) { while ( $i <= $n - 1 and $arr [ $i ] > 0) ++ $i ; while ( $j >= 0 and $arr [ $j ] < 0) -- $j ; if ( $i < $j ) { $temp = $arr [ $i ]; $arr [ $i ] = $arr [ $j ]; $arr [ $j ] = $temp ; } } // i has index of leftmost // negative element if ( $i == 0 || $i == $n ) return ; // start with first positive element // at index 0 // Rearrange array in alternating // positive & negative items $k = 0; while ( $k < $n && $i < $n ) { // swap next positive element at even // position from next negative element. $temp = $arr [ $k ]; $arr [ $k ] = $arr [ $i ]; $arr [ $i ] = $temp ; $i = $i + 1; $k = $k + 2; } } // Utility function to print an array function printArray(& $arr , $n ) { for ( $i = 0; $i < $n ; $i ++) echo $arr [ $i ] . " " ; echo "\n" ; } // Driver code $arr = array (2, 3, -4, -1, 6, -9); $n = sizeof( $arr ); echo "Given array is \n" ; printArray( $arr , $n ); rearrange( $arr , $n ); echo "Rearranged array is \n" ; printArray( $arr , $n ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to rearrange // array in alternating // positive & negative // items with O(1) extra space // Function to rearrange positive and negative // integers in alternate fashion. The below // solution doesn't maintain original order of // elements function rearrange(arr,n) { let i = 0, j = n - 1; // Shift all negative values to the end while (i < j) { while (i <= n - 1 && arr[i] > 0) i += 1; while (j >= 0 && arr[j] < 0) j -= 1; if (i < j) swap(arr, i,j); } // i has index of leftmost negative element if (i == 0 || i == n) return ; // Start with first positive // element at index 0 // Rearrange array in alternating // positive & negative items let k = 0; while (k < n && i < n) { // Swap next positive element // at even position // from next negative element. swap(arr, k, i); i = i + 1; k = k + 2; } } // Utility function to print an array function printArray(arr, n) { for (let i = 0; i < n; i++) document.write(arr[i] + " " ); document.write( "<br>" ); } function swap(arr, index1, index2) { let c = arr[index1]; arr[index1] = arr[index2]; arr[index2] = c; } // Driver code let arr = [ 2, 3, -4, -1, 6, -9 ]; let n = arr.length; document.write( "Given array is <br>" ); printArray(arr, n); rearrange(arr, n); document.write( "Rearranged array is <br>" ); printArray(arr, n); // This code is contributed by rag2127 </script> |
Given array is 2 3 -4 -1 6 -9 Rearranged array is -1 3 -4 2 -9 6
Time Complexity : O(N)
Space Complexity : O(1)