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Create a Sorted Array Using Binary Search

  • Difficulty Level : Medium
  • Last Updated : 16 Jun, 2021

Given an array, the task is to create a new sorted array in ascending order from the elements of the given array.
Examples: 
 

Input : arr[] = {2, 5, 4, 9, 8}
Output : 2 4 5 8 9

Input : arr[] = {10, 45, 98, 35, 45}
Output : 10 35 45 45 98

 

The above problem can be solved efficiently using Binary Search. We create a new array and insert the first element if it’s empty. Now for every new element, we find the correct position for the element in the new array using binary search and then insert that element at the corresponding index in the new array.
Below is the implementation of the above approach: 
 

C++




// C++ program to create a sorted array
// using Binary Search
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to create a new sorted array
// using Binary Search
void createSorted(int a[], int n)
{
    // Auxiliary Array
    vector<int> b;
 
    for (int j = 0; j < n; j++) {
        // if b is empty any element can be at
        // first place
        if (b.empty())
            b.push_back(a[j]);
        else {
 
            // Perform Binary Search to find the correct
            // position of current element in the
            // new array
            int start = 0, end = b.size() - 1;
 
            // let the element should be at first index
            int pos = 0;
 
            while (start <= end) {
 
                int mid = start + (end - start) / 2;
 
                // if a[j] is already present in the new array
                if (b[mid] == a[j]) {
                    // add a[j] at mid+1. you can add it at mid
                    b.emplace(b.begin() + max(0, mid + 1), a[j]);
                    break;
                }
                // if a[j] is lesser than b[mid] go right side
                else if (b[mid] > a[j])
                    // means pos should be between start and mid-1
                    pos = end = mid - 1;
                else
                    // else pos should be between mid+1 and end
                    pos = start = mid + 1;
 
                // if a[j] is the largest push it at last
                if (start > end) {
                    pos = start;
                    b.emplace(b.begin() + max(0, pos), a[j]);
 
                    // here max(0, pos) is used because sometimes
                    // pos can be negative as smallest duplicates
                    // can be present in the array
                    break;
                }
            }
        }
    }
 
    // Print the new generated sorted array
    for (int i = 0; i < n; i++)
        cout << b[i] << " ";
}
 
// Driver Code
int main()
{
    int a[] = { 2, 5, 4, 9, 8 };
    int n = sizeof(a) / sizeof(a[0]);
 
    createSorted(a, n);
 
    return 0;
}

Java




// Java program to create a sorted array
// using Binary Search
import java.util.*;
 
class GFG
{
 
// Function to create a new sorted array
// using Binary Search
static void createSorted(int a[], int n)
{
    // Auxiliary Array
    Vector<Integer> b = new Vector<>();
 
    for (int j = 0; j < n; j++)
    {
        // if b is empty any element can be at
        // first place
        if (b.isEmpty())
            b.add(a[j]);
        else
        {
 
            // Perform Binary Search to find the correct
            // position of current element in the
            // new array
            int start = 0, end = b.size() - 1;
 
            // let the element should be at first index
            int pos = 0;
 
            while (start <= end)
            {
 
                int mid = start + (end - start) / 2;
 
                // if a[j] is already present in the new array
                if (b.get(mid) == a[j])
                {
                    // add a[j] at mid+1. you can add it at mid
                    b.add((Math.max(0, mid + 1)), a[j]);
                    break;
                }
                 
                // if a[j] is lesser than b[mid] go right side
                else if (b.get(mid) > a[j])
                    // means pos should be between start and mid-1
                    pos = end = mid - 1;
                else
                    // else pos should be between mid+1 and end
                    pos = start = mid + 1;
 
                // if a[j] is the largest push it at last
                if (start > end)
                {
                    pos = start;
                    b.add(Math.max(0, pos), a[j]);
 
                    // here max(0, pos) is used because sometimes
                    // pos can be negative as smallest duplicates
                    // can be present in the array
                    break;
                }
            }
        }
    }
 
    // Print the new generated sorted array
    for (int i = 0; i < n; i++)
        System.out.print(b.get(i) + " ");
}
 
// Driver Code
public static void main(String args[])
{
    int a[] = { 2, 5, 4, 9, 8 };
    int n = a.length;
 
    createSorted(a, n);
}
}
 
/* This code is contributed by PrinciRaj1992 */

Python3




# Python program to create a sorted array
# using Binary Search
 
# Function to create a new sorted array
# using Binary Search
def createSorted(a: list, n: int):
 
    # Auxiliary Array
    b = []
    for j in range(n):
 
        # if b is empty any element can be at
        # first place
        if len(b) == 0:
            b.append(a[j])
        else:
 
            # Perform Binary Search to find the correct
            # position of current element in the
            # new array
            start = 0
            end = len(b) - 1
 
            # let the element should be at first index
            pos = 0
            while start <= end:
                mid = start + (end - start) // 2
 
                # if a[j] is already present in the new array
                if b[mid] == a[j]:
 
                    # add a[j] at mid+1. you can add it at mid
                    b.insert(max(0, mid + 1), a[j])
                    break
 
                # if a[j] is lesser than b[mid] go right side
                elif b[mid] > a[j]:
 
                    # means pos should be between start and mid-1
                    pos = end = mid - 1
                else:
 
                    # else pos should be between mid+1 and end
                    pos = start = mid + 1
 
                # if a[j] is the largest push it at last
                if start > end:
                    pos = start
                    b.insert(max(0, pos), a[j])
 
                    # here max(0, pos) is used because sometimes
                    # pos can be negative as smallest duplicates
                    # can be present in the array
                    break
 
    # Print the new generated sorted array
    for i in range(n):
        print(b[i], end=" ")
 
# Driver Code
if __name__ == "__main__":
 
    a = [2, 5, 4, 9, 8]
    n = len(a)
    createSorted(a, n)
 
# This code is contributed by
# sanjeev2552

C#




// C# program to create a sorted array
// using Binary Search
using System;
using System.Collections.Generic;
     
class GFG
{
 
// Function to create a new sorted array
// using Binary Search
static void createSorted(int []a, int n)
{
    // Auxiliary Array
    List<int> b = new List<int>();
 
    for (int j = 0; j < n; j++)
    {
        // if b is empty any element can be at
        // first place
        if (b.Count == 0)
            b.Add(a[j]);
        else
        {
 
            // Perform Binary Search to find the correct
            // position of current element in the
            // new array
            int start = 0, end = b.Count - 1;
 
            // let the element should be at first index
            int pos = 0;
 
            while (start <= end)
            {
 
                int mid = start + (end - start) / 2;
 
                // if a[j] is already present in the new array
                if (b[mid] == a[j])
                {
                    // add a[j] at mid+1. you can add it at mid
                    b.Insert((Math.Max(0, mid + 1)), a[j]);
                    break;
                }
                 
                // if a[j] is lesser than b[mid] go right side
                else if (b[mid] > a[j])
                 
                    // means pos should be between start and mid-1
                    pos = end = mid - 1;
                else
                 
                    // else pos should be between mid+1 and end
                    pos = start = mid + 1;
 
                // if a[j] is the largest push it at last
                if (start > end)
                {
                    pos = start;
                    b.Insert(Math.Max(0, pos), a[j]);
 
                    // here Max(0, pos) is used because sometimes
                    // pos can be negative as smallest duplicates
                    // can be present in the array
                    break;
                }
            }
        }
    }
 
    // Print the new generated sorted array
    for (int i = 0; i < n; i++)
        Console.Write(b[i] + " ");
}
 
// Driver Code
public static void Main(String []args)
{
    int []a = { 2, 5, 4, 9, 8 };
    int n = a.Length;
 
    createSorted(a, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
      // JavaScript program to create a sorted array
      // using Binary Search
       
      // Function to create a new sorted array
      // using Binary Search
      function createSorted(a, n) {
        // Auxiliary Array
        var b = [];
 
        for (var j = 0; j < n; j++) {
          // if b is empty any element can be at
          // first place
          if (b.length == 0) b.push(a[j]);
          else {
            // Perform Binary Search to find the correct
            // position of current element in the
            // new array
            var start = 0,
              end = b.length - 1;
 
            // let the element should be at first index
            var pos = 0;
 
            while (start <= end) {
              var mid = start + parseInt((end - start) / 2);
 
              // if a[j] is already present in the new array
              if (b[mid] === a[j]) {
                // add a[j] at mid+1. you can add it at mid
                b.insert(Math.max(0, mid + 1), a[j]);
 
                break;
              }
 
              // if a[j] is lesser than b[mid] go right side
              else if (b[mid] > a[j])
                // means pos should be between start and mid-1
                pos = end = mid - 1;
              // else pos should be between mid+1 and end
              else pos = start = mid + 1;
 
              // if a[j] is the largest push it at last
              if (start > end) {
                pos = start;
                b.insert(Math.max(0, pos), a[j]);
 
                // here Max(0, pos) is used because sometimes
                // pos can be negative as smallest duplicates
                // can be present in the array
                break;
              }
            }
          }
        }
 
        // Print the new generated sorted array
        for (var i = 0; i < n; i++) document.write(b[i] + " ");
      }
 
      Array.prototype.insert = function (index, item) {
        this.splice(index, 0, item);
      };
      // Driver Code
      var a = [2, 5, 4, 9, 8];
      var n = a.length;
 
      createSorted(a, n);
       
</script>
Output: 
2 4 5 8 9

 

Time Complexity: O(N*N). Although binary search is being used, the list insert calls run in O(N) time on average
Auxiliary Space: O(N)
 

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