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# C++ Program to Rearrange positive and negative numbers in O(n) time and O(1) extra space

• Last Updated : 28 May, 2022

An array contains both positive and negative numbers in random order. Rearrange the array elements so that positive and negative numbers are placed alternatively. 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 in the end of the array.
For example, if the input array is [-1, 2, -3, 4, 5, 6, -7, 8, 9], then the output should be [9, -7, 8, -3, 5, -1, 2, 4, 6]
Note: The partition process changes relative order of elements. I.e. the order of the appearance of elements is not maintained with this approach. See this for maintaining order of appearance of elements in this problem.
The solution is to first separate positive and negative numbers using partition process of QuickSort. In the partition process, consider 0 as value of 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 next positive number.

## C++

 `// A C++ program to put positive` `// numbers at even indexes (0, 2, 4,..) ` `// and negative numbers at odd ` `// indexes (1, 3, 5, ..)` `#include ` `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 elements` `void` `GFG :: swap(``int` `*a, ``int` `*b)` `{` `    ``int` `temp = *a;` `    ``*a = *b;` `    ``*b = temp;` `}`   `// A utility function to print an array` `void` `GFG :: printArray(``int` `arr[], ``int` `n)` `{` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``cout << arr[i] << ``" "``;` `}`   `// Driver Code` `int` `main() ` `{` `    ``int` `arr[] = {-1, 2, -3, 4, ` `                  ``5, 6, -7, 8, 9};` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``GFG test;` `    ``test.rearrange(arr, n);` `    ``test.printArray(arr, n);` `    ``return` `0;` `}`   `// This code is contributed ` `// by vt_Yogesh Shukla 1`

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.

Please refer complete article on Rearrange positive and negative numbers in O(n) time and O(1) extra space for more details!

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