Given a binary array, sort it using one traversal and no extra space.
Examples :
Input : 1 0 0 1 0 1 0 1 1 1 1 1 1 0 0 1 1 0 1 0 0 Output : 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 Explanation: The output is a sorted array of 0 and 1 Input : 1 0 1 0 1 0 1 0 Output : 0 0 0 0 1 1 1 1 Explanation: The output is a sorted array of 0 and 1
Approach: This concept is related to partition of quick sort . In quick sort’ partition, after one scan, the left of the array is smallest and right of the array is the largest of selected pivot element.
Algorithm:
- Create a variable index index = 0
- Traverse the array from start to end
- If the element is 0 then swap the current element with the element at index position and increment the index by 1.
- If the element is 1 keep the element as it is.
Implementation:
CPP
// CPP program to sort a binary array#include <iostream>using namespace std; void sortBinaryArray(int a[], int n){ int j = -1; for (int i = 0; i < n; i++) { // if number is smaller than 1 // then swap it with j-th number if (a[i] < 1) { j++; swap(a[i], a[j]); } }} // Driver codeint main(){ int a[] = { 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0 }; int n = sizeof(a) / sizeof(a[0]); sortBinaryArray(a, n); for (int i = 0; i < n; i++) cout << a[i] << " "; return 0;} |
Java
// JAVA Code for Sort a binary// array using one traversalimport java.util.*; class GFG { static void sortBinaryArray(int a[], int n) { int j = -1; for (int i = 0; i < n; i++) { // if number is smaller than 1 // then swap it with j-th number if (a[i] < 1) { j++; int temp = a[j]; a[j] = a[i]; a[i] = temp; } } } /* Driver program to test above function */ public static void main(String[] args) { int a[] = { 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0 }; int n = a.length; sortBinaryArray(a, n); for (int i = 0; i < n; i++) System.out.print(a[i] + " "); }} |
Python3
# A Python program to sort a# binary arraydef sortBinaryArray(a, n): j = -1 for i in range(n): # if number is smaller # than 1 then swap it # with j-th number if a[i] < 1: j = j + 1 # swap a[i], a[j] = a[j], a[i] # Driver programa = [1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0]n = len(a) sortBinaryArray(a, n) for i in range(n): print(a[i], end = " ") # This code is contributed by Shrikant13. |
C#
// C# Code for Sort a binary// array using one traversalusing System; class GFG { static void sortBinaryArray(int[] a, int n) { int j = -1; for (int i = 0; i < n; i++) { // if number is smaller than // 1 then swap it with j-th // number if (a[i] < 1) { j++; int temp = a[j]; a[j] = a[i]; a[i] = temp; } } } /* Driver program to test above function */ public static void Main() { int[] a = { 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0 }; int n = a.Length; sortBinaryArray(a, n); for (int i = 0; i < n; i++) Console.Write(a[i] + " "); }} // This code is contributed by vt_m. |
PHP
<?php// PHP Code for Sort a binary// array using one traversalfunction sortBinaryArray($a, $n){ $j = -1; for ($i = 0; $i < $n; $i++) { // if number is smaller than // 1 then swap it with j-th // number if ($a[$i] < 1) { $j++; $temp = $a[$j]; $a[$j] = $a[$i]; $a[$i] = $temp; } }for ($i = 0; $i < $n; $i++) echo $a[$i] . " "; } // Driver Code$a = array(1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0); $n = count($a);sortBinaryArray($a, $n); // This code is contributed by Sam007?> |
Output :
0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1
Complexity Analysis:
Only one traversal of the array is needed, So time Complexity is O(n).
The space required is constant.
This article is contributed by Devanshu Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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