Rearrange positive and negative numbers using inbuilt sort function

• Difficulty Level : Medium
• Last Updated : 03 Sep, 2021

Given an array of positive and negative numbers, arrange them such that all negative integers appear before all the positive integers in the array without using any additional data structure like a hash table, arrays, etc. The order of appearance should be maintained.
Examples:

Input :  arr[] = [12, 11, -13, -5, 6, -7, 5, -3, -6]
Output : arr[] = [-13, -5, -7, -3, -6, 12, 11, 6, 5]

Input :  arr[] = [-12, 11, 0, -5, 6, -7, 5, -3, -6]
Output : arr[] =  [-12, -5, -7, -3, -6, 0, 11, 6, 5]

Previous Approaches: Some approaches have already been discussed here. They were implemented at best.
Approach 3: There is another method to do so. In c++ STL, There is an inbuilt function std::sort(). We can modify the comp() function to obtain the desired result. As we have to place negative numbers first and then positive numbers. We also have to keep zero’s(if present) between positive and negative numbers.
The comp() function in this code rearranges the given array in the required order. Here in bool comp(int a, int b), if integer ‘a’ is of j-th index and integer ‘b’ is of i-th index elements in the arr[], then j>i. comp() function will be called in this way. If the comp() return true then swap will be done.

C++

 // CPP program to rearrange positive// and negative integers keeping// order of elements.#include  using namespace std; bool comp(int a, int b){ // swap not neededif((a > 0 && b > 0) ||   (a < 0 && b < 0) ||   (a > 0 && b < 0 ))return false; // swap neededif(a < 0 && b > 0)return true; // swap not neededif((a == 0 && b < 0) ||   (a > 0 && b == 0))return false; // swap neededif((a == 0 && b > 0) ||   (a < 0 && b == 0))return true; } void rearrange(int arr[], int n){    sort(arr, arr + n, comp);} // Driver codeint main(){    int arr[] = { -12, 11, -13, -5,                  6, -7, 5, -3, -6 };    int n = sizeof(arr) / sizeof(arr);    rearrange(arr, n);    for (int i = 0; i < n; i++)        cout << " " << arr[i];     return 0;}
Output:
-12 -13 -5 -7 -3 -6 11 6 5

Time complexity is the same as sorting i.e. O(n log n). As we are using the standard sort function. But it is really faster because the inbuilt sort function uses introsort.
Approach 4: There is yet another method to solve this problem. We recursively traverse the array cutting it into two halves (array[start..start] & array[(start + 1)..end], and keep on splitting the array till we reach the last element. Then we start merging it back. The idea is to, at any point, keep the array in the proper sequence of negative and positive integers. The merging logic would be:
(I) If the array[start] is negative, merge the rest of the array as it is so that the negative numbers’ order is maintained. The reason for this is that since we are tracing back from the recursive calls, we start moving right to left through the array, thus, naturally maintaining the original sequence.
(II) If the array[start] is positive, merge the rest of the array, but, after right-rotating the half of the array[(start + 1)..end]. The idea for the rotation is to merge the array so that the positive array[start] is always merged with the positive elements. But, the only thing here is that the merged array will have all the positive elements on the left and negative elements on the right. So we reverse the sequence in each recursion to get back the original sequence of negative elements and then positive elements subsequently.
It can be observed since we reverse the array while merging with a positive first element in each recursion, so the sequence of positive elements, although coming after the negative elements, are in reverse order. So, as a final step, we reverse only the positive half of the final array, and, subsequently getting the intended sequence.
Below is the implementation of the above approach:

C++

 // C++ implementation of// the above approach#include  void printArray(int array[], int length){    std::cout << "[";         for(int i = 0; i < length; i++)    {        std::cout << array[i];                 if(i < (length - 1))            std::cout << ", ";        else            std::cout << "]" << std::endl;    }} void reverse(int array[], int start, int end){    while(start < end)    {        int temp = array[start];        array[start] = array[end];        array[end] = temp;        start++;        end--;    }} // Rearrange the array with all negative integers// on left and positive integers on right// use recursion to split the array with first element// as one half and the rest array as another and then// merge it with head of the array in each step void rearrange(int array[], int start, int end){    // exit condition    if(start == end)        return;         // rearrange the array except the first    // element in each recursive call    rearrange(array, (start + 1), end);         // If the first element of the array is positive,    // then right-rotate the array by one place first    // and then reverse the merged array.    if(array[start] >= 0)    {        reverse(array, (start + 1), end);        reverse(array, start, end);    }} // Driver codeint main(){    int array[] = {-12, -11, -13, -5, -6, 7, 5, 3, 6};    int length = (sizeof(array) / sizeof(array));    int countNegative = 0;         for(int i = 0; i < length; i++)    {        if(array[i] < 0)            countNegative++;    }         std::cout << "array: ";    printArray(array, length);    rearrange(array, 0, (length - 1));         reverse(array, countNegative, (length - 1));         std::cout << "rearranged array: ";    printArray(array, length);    return 0;}

Java

 // Java program to implement the// above approachimport java.io.*;class GFG{ static void printArray(int[] array, int length){  System.out.print("[");   for (int i = 0; i < length; i++)  {    System.out.print(array[i]);     if (i < (length - 1))      System.out.print(",");    else      System.out.print("]\n");  }} static void reverse(int[] array,                    int start,                    int end){  while (start < end)  {    int temp = array[start];    array[start] = array[end];    array[end] = temp;    start++;    end--;  }} // Rearrange the array with// all negative integers on left// and positive integers on right// use recursion to split the// array with first element// as one half and the rest// array as another and then// merge it with head of// the array in each stepstatic void rearrange(int[] array,                      int start,                      int end){  // exit condition  if (start == end)    return;   // rearrange the array  // except the first element  // in each recursive call  rearrange(array,            (start + 1), end);   // If the first element of  // the array is positive,  // then right-rotate the  // array by one place first  // and then reverse the merged array.  if (array[start] >= 0)  {    reverse(array,            (start + 1), end);    reverse(array,            start, end);  }} // Driver codepublic static void main(String[] args){   int[] array = {-12, -11, -13,                 -5, -6, 7, 5, 3, 6};  int length = array.length;  int countNegative = 0;   for (int i = 0; i < length; i++)  {    if (array[i] < 0)      countNegative++;  }   System.out.print("array: ");  printArray(array, length);  rearrange(array, 0,           (length - 1));  reverse(array, countNegative,         (length - 1));  System.out.print("rearranged array: ");  printArray(array, length);}} // This code is contributed by Chitranayal

Python3

 # Python3 implementation of the above approachdef printArray(array, length):    print("[", end = "")     for i in range(length):        print(array[i], end = "")         if(i < (length - 1)):            print(",", end = " ")        else:            print("]") def reverse(array, start, end):    while(start < end):        temp = array[start]        array[start] = array[end]        array[end] = temp        start += 1        end -= 1 # Rearrange the array with all negative integers# on left and positive integers on right# use recursion to split the array with first element# as one half and the rest array as another and then# merge it with head of the array in each stepdef rearrange(array, start, end):         # exit condition    if(start == end):        return     # rearrange the array except the first    # element in each recursive call    rearrange(array, (start + 1), end)     # If the first element of the array is positive,    # then right-rotate the array by one place first    # and then reverse the merged array.    if(array[start] >= 0):        reverse(array, (start + 1), end)        reverse(array, start, end) # Driver codeif __name__ == '__main__':    array = [-12, -11, -13, -5, -6, 7, 5, 3, 6]    length = len(array)    countNegative = 0     for i in range(length):        if(array[i] < 0):            countNegative += 1     print("array: ", end = "")    printArray(array, length)    rearrange(array, 0, (length - 1))     reverse(array, countNegative, (length - 1))     print("rearranged array: ", end = "")    printArray(array, length) # This code is contributed by mohit kumar 29

C#

 // C# implementation of// the above approachusing System;class GFG{ static void printArray(int []array,                       int length){  Console.Write("[");   for(int i = 0; i < length; i++)  {    Console.Write(array[i]);     if(i < (length - 1))      Console.Write(",");    else      Console.Write("]\n");  }}  static void reverse(int []array,                    int start, int end){  while(start < end)  {    int temp = array[start];    array[start] = array[end];    array[end] = temp;    start++;    end--;  }}  // Rearrange the array with// all negative integers on left// and positive integers on right// use recursion to split the// array with first element// as one half and the rest// array as another and then// merge it with head of// the array in each step static void rearrange(int []array,                      int start, int end){  // exit condition  if(start == end)    return;   // rearrange the array  // except the first element  // in each recursive call  rearrange(array,            (start + 1), end);   // If the first element of  // the array is positive,  // then right-rotate the  // array by one place first  // and then reverse the merged array.  if(array[start] >= 0)  {    reverse(array, (start + 1), end);    reverse(array, start, end);  }}      // Driver codepublic static void Main(string[] args){   int []array = {-12, -11, -13,                 -5, -6, 7, 5, 3, 6};  int length = array.Length;  int countNegative = 0;   for(int i = 0; i < length; i++)  {    if(array[i] < 0)      countNegative++;  }     Console.Write("array: ");  printArray(array, length);  rearrange(array, 0, (length - 1));  reverse(array, countNegative, (length - 1));  Console.Write("rearranged array: ");  printArray(array, length);}} // This code is contributed by Rutvik_56

Javascript


Output:
array: [-12, -11, -13, -5, -6, 7, 5, 3, 6]
rearranged array: [-12, -11, -13, -5, -6, 7, 5, 3, 6]

Time complexity: O(N^2)
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