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 1Input: 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
Sort a binary array using one traversal using partition function of quicksort:
This concept is related to partition of quick sort . In the quick sort partition function, after one scan, the left of the array is the smallest and the right of the array is the largest of the selected pivot element
Follow the given steps to solve the problem:
- Create a variable index say j = -1
- Traverse the array from start to end
- If the element is 0 then swap the current element with the element at the index( jth ) position and increment the index j by 1.
- If the element is 1 keep the element as it is.
Below is the implementation of the above approach:
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 code
int 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]);
// Function call
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 traversal
import 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 code
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;
// Function call
sortBinaryArray(a, n);
for (int i = 0; i < n; i++)
System.out.print(a[i] + " ");
}
}
Python3
# A Python program to sort a
# binary array
def 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 code
if __name__ == "__main__":
a = [1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1,
1, 1, 0, 0, 1, 1, 0, 1, 0, 0]
n = len(a)
# Function call
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 traversal
using 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 code
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;
// Function call
sortBinaryArray(a, n);
for (int i = 0; i < n; i++)
Console.Write(a[i] + " ");
}
}
// This code is contributed by vt_m.
Javascript
// Javascript Code for Sort a binary
// array using one traversal
function sortBinaryArray(a, n)
{
let j = -1;
for (let i = 0; i < n; i++) {
// if number is smaller than 1
// then swap it with j-th number
if (a[i] < 1) {
j++;
let temp = a[j];
a[j] = a[i];
a[i] = temp;
}
}
}
// driver function
let a = [ 1, 0, 0, 1, 0, 1, 0, 1, 1, 1,
1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0 ];
let n = a.length;
sortBinaryArray(a, n);
for (let i = 0; i < n; i++)
console.log(a[i] + " ");
// This code is contributed by code_hunt.
PHP
<?php
// PHP Code for Sort a binary
// array using one traversal
function 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);
// Function call
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
Time Complexity: O(N), Only one traversal of the array is needed, So the time Complexity is O(N)
Auxiliary Space: O(1). The space required is constant
Sort a binary array using one traversal and no extra space using Two Pointer
Step-by-step approach:
- The function initializes two pointers i and j at the start and end of the array respectively (i at 0 and j at N-1).
- Inside the loop, it checks if the element at index i is 1 and the element at index j is 0. If this condition is true, it swaps the elements. Then it increments i and decrements j.
- If the element at index i is 0, it increments i. This ensures that i moves forward until it finds a 1.
- If the element at index j is 1, it decrements j. This ensures that j moves backward until it finds a 0.
Below is the implementation of the above Approach:
C++
#include <iostream>
#include <vector>
void solve(std::vector<int>& arr, int n) {
// Initialize two pointers, one at the beginning and one at the end of the array.
int i = 0;
int j = n - 1;
// Iterate until the two pointers meet.
while (i < j) {
// If the element at the current index is 1 and the element at the end of the array is 0,
// swap the two elements.
if (arr[i] == 1 && arr[j] == 0) {
std::swap(arr[i], arr[j]);
// Move the left pointer one position to the right.
i++;
// Move the right pointer one position to the left.
j--;
}
// If the element at the current index is 0, move the left pointer one position to the right.
else if (arr[i] == 0) {
i++;
}
// If the element at the end of the array is 1, move the right pointer one position to the left.
else if (arr[j] == 1) {
j--;
}
}
}
int main() {
// Example usage
std::vector<int> arr = {0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1};
int n = arr.size();
solve(arr, n);
// Output the modified array
for (int num : arr) {
std::cout << num << " ";
}
std::cout << std::endl; // Output: 0 0 0 0 0 1 1 1 1 1 1
return 0;
}
Java
import java.io.*;
class GFG {
public static void Solve(int A[], int n)
{
// Initialize two pointers, one at the beginning and
// one at the end of the array.
int i = 0;
int j = n - 1;
// Iterate until the two pointers meet.
while (i < j) {
// If the element at the current index is 1 and
// the element at the end of the array is 0,
// swap the two elements.
if (A[i] == 1 && A[j] == 0) {
int temp = A[i];
A[i] = A[j];
A[j] = temp;
// Move the left pointer one position to the
// right.
i++;
// Move the right pointer one position to
// the left.
j--;
}
// If the element at the current index is 0,
// move the left pointer one position to the
// right.
else if (A[i] == 0) {
i++;
}
// If the element at the end of the array is 1,
// move the right pointer one position to the
// left.
else if (A[j] == 1) {
j--;
}
}
}
public static void main(String[] args)
{
// Create an array of 0s and 1s.
int arr[] = { 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1 };
// Sort the array using the Solve() method.
Solve(arr, arr.length);
// Print the sorted array.
for (int i : arr) {
System.out.print(i + " ");
}
}
}
Python3
def solve(arr, n):
# Initialize two pointers, one at the beginning and one at the end of the array.
i = 0
j = n - 1
# Iterate until the two pointers meet.
while i < j:
# If the element at the current index is 1 and the element at the end of the array is 0,
# swap the two elements.
if arr[i] == 1 and arr[j] == 0:
arr[i], arr[j] = arr[j], arr[i]
# Move the left pointer one position to the right.
i += 1
# Move the right pointer one position to the left.
j -= 1
# If the element at the current index is 0, move the left pointer one position to the right.
elif arr[i] == 0:
i += 1
# If the element at the end of the array is 1, move the right pointer one position to the left.
elif arr[j] == 1:
j -= 1
# Example usage
arr = [0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1]
solve(arr, len(arr))
print(arr) # Output: [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]
JavaScript
function solve(arr) {
// Initialize two pointers, one at the beginning and one at the end of the array.
let i = 0;
let j = arr.length - 1;
// Iterate until the two pointers meet.
while (i < j) {
// If the element at the current index is 1 and the element at the end of the array is 0,
// swap the two elements.
if (arr[i] === 1 && arr[j] === 0) {
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
// Move the left pointer one position to the right.
i++;
// Move the right pointer one position to the left.
j--;
}
// If the element at the current index is 0, move the left pointer one position to the right.
else if (arr[i] === 0) {
i++;
}
// If the element at the end of the array is 1, move the right pointer one position to the left.
else if (arr[j] === 1) {
j--;
}
}
}
// Example usage
let arr = [0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1];
solve(arr);
console.log("Sorted array:", arr);
Output
0 0 0 0 0 1 1 1 1 1 1
Time Complexity: O(N)
Auxiliary Space: O(1)