Sort an array by swapping adjacent elements from indices that contains '1' in a given string

Given an array arr[] of size N and a binary string S, the task is to check if it is possible to sort the array arr[] by swapping adjacent array elements, say arr[i] and arr[i + 1] if S[i] is equal to '1'. If it is possible, then print "Yes". Otherwise, print "No".

Examples :

Input: N = 6, arr[] = {2, 5, 3, 4, 6}, S = "01110"
Output: Yes
Explanation: 
Indices that can be swapped are {1, 2, 3}.
Swapping arr[1] and arr[2] modifies the array arr[] to {1, 2, 3, 5, 4, 6}. 
Swapping arr[3] and arr[4] modifies the array arr[] to {1, 2, 3, 4, 5, 6}.

Input : N = 6, arr[] = {1, 2, 5, 3, 4, 6}, S = "01010"
Output : No

Approach: Take a look at some pair (i, ?j) in the array arr[] such that i?<?j and initial arr[i]?>?arr[j]. That means that all the swaps from i to j?-?1 should be allowed. Then it's easy to notice that it's enough to check only i and i?+?1 as any other pair can be deducted from this. Follow the steps below to solve the problem:

• Iterate over the range [0, N - 1] and perform the following steps:
  • If S[i] is '1', then initialize a variable, say j, as equal to i.
  • Iterate over the range till s[j] is equal to '1' and j is less than the length of the string s. Increase the value of j by 1.
  • Sort the subarray in the array arr[] from i to j+1.
• Iterate over the range [0, N - 2] and perform the following steps:
  • If the value of arr[i] is greater than arr[i + 1], then print No and break the loop and return.
• Print Yes as the array is sorted by swapping the allowed indices.

Below is the implementation of the above approach.

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if it is possible
// to sort the array arr[] by swapping
// array elements from indices containing
// adjacent pairs of 1s in the string s
void checkIfPossibleToSort(int n, int arr[],
                           string s)
{
    // Sort the parts of array
    // where swaps are allowed
    for (int i = 0; i < n - 1; i++) {

        if (s[i] == '1') {
            int j = i;

            // Iterate over the range
            // till s[j] is equal to '1'
            while (s[j] == '1') {
                j++;
            }

            // Sort the subarray
            sort(arr + i, arr + j + 1);

            i = j;
        }
    }

    // Check if the array remains unsorted
    for (int i = 0; i < n - 1; i++) {

        // If found to be true, then it is
        // not possible to sort the array
        if (arr[i] > arr[i + 1]) {
            printf("No\n");
            return;
        }
    }

    printf("Yes\n");
}

// Driver Code
int main()
{
    // Given Input
    int n = 6;
    int arr[] = { 2, 5, 3, 4, 6 };
    string s = "01110";

    // Function Call
    checkIfPossibleToSort(n, arr, s);

    return 0;
}

// Java program for the above approach
import java.util.Arrays;

class GFG{

// Function to check if it is possible
// to sort the array arr[] by swapping
// array elements from indices containing
// adjacent pairs of 1s in the string s
public static void checkIfPossibleToSort(int n, int arr[], 
                                         String s) 
{
    
    // Sort the parts of array
    // where swaps are allowed
    for(int i = 0; i < n - 1; i++)
    {
        if (s.charAt(i) == '1') 
        {
            int j = i;

            // Iterate over the range
            // till s[j] is equal to '1'
            while (s.charAt(j) == '1')
            {
                j++;
            }

            // Sort the subarray
            Arrays.sort(arr, i, j);

            i = j;
        }
    }

    // Check if the array remains unsorted
    for(int i = 0; i < n - 2; i++)
    {
        
        // If found to be true, then it is
        // not possible to sort the array
        if (arr[i] > arr[i + 1]) 
        {
            System.out.println("No");
            return;
        }
    }
    System.out.println("Yes");
}

// Driver Code
public static void main(String args[]) 
{
    
    // Given Input
    int n = 6;
    int arr[] = { 2, 5, 3, 4, 6 };
    String s = "01110";

    // Function Call
    checkIfPossibleToSort(n, arr, s);
}
}

// This code is contributed by gfgking

# Python3 program for the above approach

# Function to check if it is possible
# to sort the array arr[] by swapping
# array elements from indices containing
# adjacent pairs of 1s in the string s
def checkIfPossibleToSort(n, arr, s):
  
   # Sort the parts of array
    # where swaps are allowed
    for i in range(n-1):
        if s[i] == '1':
            j = i
            
            # Iterate over the range
            # till s[j] is equal to '1'
            while s[j] == '1':
                j += 1
                
                # sort the array
            arr = arr[:i] + sorted(arr[i:j+1]) + arr[j+1:]
            i = j
            
     # Check if the array remains unsorted
    for i in range(n-2):
      
      # If found to be true, then it is
       # not possible to sort the array
        if arr[i] > arr[i+1]:
            print("No")
            return
    print("Yes")


    # Driver code
    
    # Given input
n = 6
arr = [2, 5, 3, 4, 6]
s = "01110"

# function call
checkIfPossibleToSort(n, arr, s)

# This code is contributed by Parth Manchanda

// C# program for the above approach
using System;

class GFG{

// Function to check if it is possible
// to sort the array arr[] by swapping
// array elements from indices containing
// adjacent pairs of 1s in the string s
public static void checkIfPossibleToSort(int n, int []arr, String s) 
{
    
    // Sort the parts of array
    // where swaps are allowed
    for(int i = 0; i < n - 1; i++)
    {
        if (s[i] == '1') 
        {
            int j = i;

            // Iterate over the range
            // till s[j] is equal to '1'
            while (s[j] == '1')
            {
                j++;
            }

            // Sort the subarray
            Array.Sort(arr, i, j);

            i = j;
        }
    }

    // Check if the array remains unsorted
    for(int i = 0; i < n - 2; i++)
    {
        
        // If found to be true, then it is
        // not possible to sort the array
        if (arr[i] > arr[i + 1]) 
        {
            Console.Write("No");
            return;
        }
    }
    Console.Write("Yes");
}

// Driver Code
public static void Main(String []args) 
{
    
    // Given Input
    int n = 6;
    int []arr = { 2, 5, 3, 4, 6 };
    String s = "01110";

    // Function Call
    checkIfPossibleToSort(n, arr, s);
}
}

// This code is contributed by shivanisinghss2110

<script>
// Javascript program for the above approach

// Function to check if it is possible
// to sort the array arr[] by swapping
// array elements from indices containing
// adjacent pairs of 1s in the string s
function checkIfPossibleToSort(n, arr, s) {
  // Sort the parts of array
  // where swaps are allowed
  for (let i = 0; i < n - 1; i++) {
    if (s[i] == "1") {
      let j = i;

      // Iterate over the range
      // till s[j] is equal to '1'
      while (s[j] == "1") {
        j++;
      }

      // Sort the subarray
      arr = [
        ...arr.slice(0, i),
        ...arr.slice(i, j + 1).sort((a, b) => a - b),
        ...arr.slice(j + 1, arr.length - 1),
      ];

      i = j;
    }
  }

  // Check if the array remains unsorted
  for (let i = 0; i < n - 1; i++) {
    // If found to be true, then it is
    // not possible to sort the array
    if (arr[i] > arr[i + 1]) {
      document.write("No<br>");
      return;
    }
  }

  document.write("Yes\n");
}

// Driver Code

// Given Input
let n = 6;
let arr = [2, 5, 3, 4, 6];
let s = "01110";

// Function Call
checkIfPossibleToSort(n, arr, s);

// tHIS CODE IS CONTRIBUTED BY _SAURABH_JAISWAL.
</script>

Yes

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)

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Article Tags :

• Strings
• Sorting
• Competitive Programming
• DSA
• Arrays
• array-rearrange

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Practice Tags :

• Arrays
• Sorting
• Strings