Given an array having positive and negative numbers, our task is to arrange them in an alternate fashion such that every positive number is followed by a negative number and vice-versa maintaining the order of appearance. The number of positive and negative numbers need not to be equal. If there are more positive numbers then they have to appear at the end of the array , same condition for negative numbers also .
Examples:
Input: arr[] = {1, 2, 3, -4, -1, 4}
Output: arr[] = {-4, 1, -1, 2, 3, 4}Input: arr[] = {-5, -2, 5, 2, 4, 7, 1, 8, 0, -8}
Output: arr[] = {-5, 5, -2, 2, -8, 4, 7, 1, 8, 0}
This question has been asked in many places (See this and this)
Naïve Approach: To solve the problem follow the below idea:
- The problem can be easily solved if O(n) extra space is allowed.
- We can store the positive values and negative values in two separate data structures.
- We will start filling the original array with alternating negative and positive values in the same order
in which it appears in the original array.
It becomes interesting due to the limitations that O(1) extra space and order of appearances.
Optimal Approach: To solve the problem in O(1) Auxiliary space follow the below idea:
The idea is to process the array from left to right. While processing, find the first out-of-place element in the remaining unprocessed array. An element is out of place if it is negative and at odd index (0-based index), or if it is positive and at even index (0-based index). Once we find an out-of-place element, we find the first element after it with an opposite sign. We right rotate the subarray between these two elements (including these two).
Illustration:
Let the array be arr[] = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 }
First iteration:
- { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 } , -2 appears on odd index position and is out of place.
We will look for the first element that appears with an opposite sign- { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 } => perform rotation of subarray between this two elements
- { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 }
Second iteration:
- { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 } , 4 is out of place.
- { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 } => -8 is of different sign
- { -5, 5, -2, 2, -8, 4, 7, 1, 8, 0 } => after performing right rotation between 4 to -8
resultant array arr[] = { -5, 5, -2, 2, -8, 4, 7, 1, 8, 0 }
Follow the steps to solve the problem:
- Maintain a variable to mark if the element is in its correct position or not. Let the variable be outofplace. Initially, it is -1.
- We will iterate over the array.
- If outofplace is -1, we will check if the current index is out of place.
- If the current index is out of place we will update the outofplace with the index value.
- Else we will keep the value as it is.
- If the outofplace is not -1, we will search for the next index which has a different sign than that of the value that is present in outofplace position.
- Now we will pass this two positions to right rotate our array.
- Update the value of outofplace by 2 unit (becuase previously valid elements will now become the out-of-place elements).
Following is the implementation of the above idea.
C++
/* C++ program to rearrange positive and negative integers in alternate fashion while keeping the order of positive and negative numbers. */#include <bits/stdc++.h>using namespace std;// Utility function to right rotate all elements between// [outofplace, cur]void rightrotate(int arr[], int n, int outofplace, int cur){ char tmp = arr[cur]; for (int i = cur; i > outofplace; i--) arr[i] = arr[i - 1]; arr[outofplace] = tmp;}void rearrange(int arr[], int n){ int outofplace = -1; for (int index = 0; index < n; index++) { if (outofplace >= 0) { // find the item which must be moved into the // out-of-place entry if out-of-place entry is // positive and current entry is negative OR if // out-of-place entry is negative and current // entry is positive then right rotate // // [...-3, -4, -5, 6...] --> [...6, -3, -4, // -5...] // ^ ^ // | | // outofplace --> outofplace // if (((arr[index] >= 0) && (arr[outofplace] < 0)) || ((arr[index] < 0) && (arr[outofplace] >= 0))) { rightrotate(arr, n, outofplace, index); // the new out-of-place entry is now 2 steps // ahead if (index - outofplace >= 2) outofplace = outofplace + 2; else outofplace = -1; } } // if no entry has been flagged out-of-place if (outofplace == -1) { // check if current entry is out-of-place if (((arr[index] >= 0) && (!(index & 0x01))) || ((arr[index] < 0) && (index & 0x01))) { outofplace = index; } } }}// A utility function to print an array 'arr[]' of size 'n'void printArray(int arr[], int n){ for (int i = 0; i < n; i++) cout << arr[i] << " "; cout << endl;}// Driver codeint main(){ int arr[] = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Given array is \n"; printArray(arr, n); // Function Call rearrange(arr, n); cout << "Rearranged array is \n"; printArray(arr, n); return 0;} |
Java
import java.io.*;class RearrangeArray { // Utility function to right rotate all elements // between [outofplace, cur] void rightrotate(int arr[], int n, int outofplace, int cur) { int tmp = arr[cur]; for (int i = cur; i > outofplace; i--) arr[i] = arr[i - 1]; arr[outofplace] = tmp; } void rearrange(int arr[], int n) { int outofplace = -1; for (int index = 0; index < n; index++) { if (outofplace >= 0) { // find the item which must be moved into // the out-of-place entry if out-of-place // entry is positive and current entry is // negative OR if out-of-place entry is // negative and current entry is negative // then right rotate // // [...-3, -4, -5, 6...] --> [...6, -3, // -4, -5...] // ^ ^ // | | // outofplace --> outofplace // if (((arr[index] >= 0) && (arr[outofplace] < 0)) || ((arr[index] < 0) && (arr[outofplace] >= 0))) { rightrotate(arr, n, outofplace, index); // the new out-of-place entry is now 2 // steps ahead if (index - outofplace >= 2) outofplace = outofplace + 2; else outofplace = -1; } } // if no entry has been flagged out-of-place if (outofplace == -1) { // check if current entry is out-of-place if (((arr[index] >= 0) && ((index & 0x01) == 0)) || ((arr[index] < 0) && (index & 0x01) == 1)) outofplace = index; } } } // A utility function to print // an array 'arr[]' of size 'n' void printArray(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); System.out.println(""); } // Driver Code public static void main(String[] args) { RearrangeArray rearrange = new RearrangeArray(); /* int arr[n] = {-5, 3, 4, 5, -6, -2, 8, 9, -1, -4}; int arr[] = {-5, -3, -4, -5, -6, 2 , 8, 9, 1 , 4}; int arr[] = {5, 3, 4, 2, 1, -2 , -8, -9, -1 , -4}; int arr[] = {-5, 3, -4, -7, -1, -2 , -8, -9, 1 , -4};*/ int arr[] = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 }; int n = arr.length; System.out.println("Given array is "); rearrange.printArray(arr, n); rearrange.rearrange(arr, n); System.out.println("RearrangeD array is "); rearrange.printArray(arr, n); }}// This code has been contributed by Mayank Jaiswal |
Python
# Python3 program to rearrange# positive and negative integers# in alternate fashion and# maintaining the order of positive# and negative numbers# rotates the array to right by once# from index 'outOfPlace to cur'def rightRotate(arr, n, outOfPlace, cur): temp = arr[cur] for i in range(cur, outOfPlace, -1): arr[i] = arr[i - 1] arr[outOfPlace] = temp return arrdef rearrange(arr, n): outOfPlace = -1 for index in range(n): if(outOfPlace >= 0): # if element at outOfPlace place in # negative and if element at index # is positive we can rotate the # array to right or if element # at outOfPlace place in positive and # if element at index is negative we # can rotate the array to right if((arr[index] >= 0 and arr[outOfPlace] < 0) or (arr[index] < 0 and arr[outOfPlace] >= 0)): arr = rightRotate(arr, n, outOfPlace, index) if(index-outOfPlace > 2): outOfPlace += 2 else: outOfPlace = - 1 if(outOfPlace == -1): # conditions for A[index] to # be in out of place if((arr[index] >= 0 and index % 2 == 0) or (arr[index] < 0 and index % 2 == 1)): outOfPlace = index return arr# Driver Codearr = [-5, -2, 5, 2, 4, 7, 1, 8, 0, -8]print("Given Array is:")print(arr)print("\nRearranged array is:")print(rearrange(arr, len(arr)))# This code is contributed# by Charan Sai |
C#
// Rearrange array in alternating positive// & negative items with O(1) extra spaceusing System;class GFG { // Utility function to right rotate // all elements between [outofplace, cur] static void rightrotate(int[] arr, int n, int outofplace, int cur) { int tmp = arr[cur]; for (int i = cur; i > outofplace; i--) arr[i] = arr[i - 1]; arr[outofplace] = tmp; } static void rearrange(int[] arr, int n) { int outofplace = -1; for (int index = 0; index < n; index++) { if (outofplace >= 0) { // find the item which must be moved // into the out-of-place entry if out-of- // place entry is positive and current // entry is negative OR if out-of-place // entry is negative and current entry // is negative then right rotate // [...-3, -4, -5, 6...] --> [...6, -3, -4, // -5...] // ^ ^ // | | // outofplace --> outofplace // if (((arr[index] >= 0) && (arr[outofplace] < 0)) || ((arr[index] < 0) && (arr[outofplace] >= 0))) { rightrotate(arr, n, outofplace, index); // the new out-of-place entry // is now 2 steps ahead if (index - outofplace > 2) outofplace = outofplace + 2; else outofplace = -1; } } // if no entry has been flagged out-of-place if (outofplace == -1) { // check if current entry is out-of-place if (((arr[index] >= 0) && ((index & 0x01) == 0)) || ((arr[index] < 0) && (index & 0x01) == 1)) outofplace = index; } } } // A utility function to print an // array 'arr[]' of size 'n' static void printArray(int[] arr, int n) { for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); Console.WriteLine(""); } // Driver code public static void Main() { int[] arr = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 }; 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 Sam007 |
Javascript
<script> // Utility function to right rotate all elements // between [outofplace, cur] function rightrotate(arr , n , outofplace , cur) { var tmp = arr[cur]; for (i = cur; i > outofplace; i--) arr[i] = arr[i - 1]; arr[outofplace] = tmp; } function rearrange(arr , n) { var outofplace = -1; for (var index = 0; index < n; index++) { if (outofplace >= 0) { // find the item which must be moved into // the out-of-place entry if out-of-place // entry is positive and current entry is // negative OR if out-of-place entry is // negative and current entry is negative // then right rotate // // [...-3, -4, -5, 6...] --> [...6, -3, // -4, -5...] // ^ ^ // | | // outofplace --> outofplace // if (((arr[index] >= 0) && (arr[outofplace] < 0)) || ((arr[index] < 0) && (arr[outofplace] >= 0))) { rightrotate(arr, n, outofplace, index); // the new out-of-place entry is now 2 // steps ahead if (index - outofplace >= 2) outofplace = outofplace + 2; else outofplace = -1; } } // if no entry has been flagged out-of-place if (outofplace == -1) { // check if current entry is out-of-place if (((arr[index] >= 0) && ((index & 0x01) == 0)) || ((arr[index] < 0) && (index & 0x01) == 1)) outofplace = index; } } } // A utility function to print // an array 'arr' of size 'n' function printArray(arr , n) { for (i = 0; i < n; i++) document.write(arr[i] + " "); document.write(""); } // Driver Code /* * var arr[n] = [-5, 3, 4, 5, -6, -2, 8, 9, -1, -4]; var arr = [-5, -3, -4, * -5, -6, 2 , 8, 9, 1 , 4]; var arr = [5, 3, 4, 2, 1, -2 , -8, -9, -1 , -4]; * var arr = [-5, 3, -4, -7, -1, -2 , -8, -9, 1 , -4]; */ var arr = [ -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 ]; var n = arr.length; document.write("Given array is "); printArray(arr, n); rearrange(arr, n); document.write("<br/>Rearranged array is "); printArray(arr, n); // This code is contributed by gauravrajput1 </script> |
PHP
<?php // PHP program to rearrange positive and // negative integers in alternate fashion // while keeping the order of positive // and negative numbers. // Utility function to right rotate all// elements between [outofplace, cur]function rightrotate(&$arr, $n, $outofplace, $cur){ $tmp = $arr[$cur]; for ($i = $cur; $i > $outofplace; $i--) $arr[$i] = $arr[$i - 1]; $arr[$outofplace] = $tmp;}function rearrange(&$arr, $n){ $outofplace = -1; for ($index = 0; $index < $n; $index ++) { if ($outofplace >= 0) { // find the item which must be moved // into the out-of-place entry if // out-of-place entry is positive and // current entry is negative OR if // out-of-place entry is negative // and current entry is negative then // right rotate // [...-3, -4, -5, 6...] --> [...6, -3, -4, -5...] // ^ ^ // | | // outofplace --> outofplace // if ((($arr[$index] >= 0) && ($arr[$outofplace] < 0)) || (($arr[$index] < 0) && ($arr[$outofplace] >= 0))) { rightrotate($arr, $n, $outofplace, $index); // the new out-of-place entry is // now 2 steps ahead if ($index - $outofplace > 2) $outofplace = $outofplace + 2; else $outofplace = -1; } } // if no entry has been flagged out-of-place if ($outofplace == -1) { // check if current entry is out-of-place if ((($arr[$index] >= 0) && (!($index & 0x01))) || (($arr[$index] < 0) && ($index & 0x01))) { $outofplace = $index; } } }}// A utility function to print an// array 'arr[]' of size 'n'function printArray(&$arr, $n){ for ($i = 0; $i < $n; $i++) echo $arr[$i]." "; echo "\n";}// Driver Code// arr = array(-5, 3, 4, 5, -6, -2, 8, 9, -1, -4);// arr = array(-5, -3, -4, -5, -6, 2 , 8, 9, 1 , 4);// arr = array(5, 3, 4, 2, 1, -2 , -8, -9, -1 , -4);// arr = array(-5, 3, -4, -7, -1, -2 , -8, -9, 1 , -4);$arr = array(-5, -2, 5, 2, 4, 7, 1, 8, 0, -8);$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?> |
Output
Given array is -5 -2 5 2 4 7 1 8 0 -8 Rearranged array is -5 5 -2 2 -8 4 7 1 8 0
Time Complexity: O(N^2), as we are using a loop to traverse N times and calling function to right rotate each time which will cost O (N).
Auxiliary Space: O(1).
Related article:
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
