Sort an array by swapping adjacent elements from indices that contains ‘1’ in a given string
Last Updated :
30 Jul, 2021
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++
#include <bits/stdc++.h>
using namespace std;
void checkIfPossibleToSort(int n, int arr[],
string s)
{
for (int i = 0; i < n - 1; i++) {
if (s[i] == '1') {
int j = i;
while (s[j] == '1') {
j++;
}
sort(arr + i, arr + j + 1);
i = j;
}
}
for (int i = 0; i < n - 1; i++) {
if (arr[i] > arr[i + 1]) {
printf("No\n");
return;
}
}
printf("Yes\n");
}
int main()
{
int n = 6;
int arr[] = { 2, 5, 3, 4, 6 };
string s = "01110";
checkIfPossibleToSort(n, arr, s);
return 0;
}
|
Java
import java.util.Arrays;
class GFG{
public static void checkIfPossibleToSort(int n, int arr[],
String s)
{
for(int i = 0; i < n - 1; i++)
{
if (s.charAt(i) == '1')
{
int j = i;
while (s.charAt(j) == '1')
{
j++;
}
Arrays.sort(arr, i, j);
i = j;
}
}
for(int i = 0; i < n - 2; i++)
{
if (arr[i] > arr[i + 1])
{
System.out.println("No");
return;
}
}
System.out.println("Yes");
}
public static void main(String args[])
{
int n = 6;
int arr[] = { 2, 5, 3, 4, 6 };
String s = "01110";
checkIfPossibleToSort(n, arr, s);
}
}
|
Python3
def checkIfPossibleToSort(n, arr, s):
for i in range(n-1):
if s[i] == '1':
j = i
while s[j] == '1':
j += 1
arr = arr[:i] + sorted(arr[i:j+1]) + arr[j+1:]
i = j
for i in range(n-2):
if arr[i] > arr[i+1]:
print("No")
return
print("Yes")
n = 6
arr = [2, 5, 3, 4, 6]
s = "01110"
checkIfPossibleToSort(n, arr, s)
|
C#
using System;
class GFG{
public static void checkIfPossibleToSort(int n, int []arr, String s)
{
for(int i = 0; i < n - 1; i++)
{
if (s[i] == '1')
{
int j = i;
while (s[j] == '1')
{
j++;
}
Array.Sort(arr, i, j);
i = j;
}
}
for(int i = 0; i < n - 2; i++)
{
if (arr[i] > arr[i + 1])
{
Console.Write("No");
return;
}
}
Console.Write("Yes");
}
public static void Main(String []args)
{
int n = 6;
int []arr = { 2, 5, 3, 4, 6 };
String s = "01110";
checkIfPossibleToSort(n, arr, s);
}
}
|
Javascript
<script>
function checkIfPossibleToSort(n, arr, s) {
for (let i = 0; i < n - 1; i++) {
if (s[i] == "1") {
let j = i;
while (s[j] == "1") {
j++;
}
arr = [
...arr.slice(0, i),
...arr.slice(i, j + 1).sort((a, b) => a - b),
...arr.slice(j + 1, arr.length - 1),
];
i = j;
}
}
for (let i = 0; i < n - 1; i++) {
if (arr[i] > arr[i + 1]) {
document.write("No<br>");
return;
}
}
document.write("Yes\n");
}
let n = 6;
let arr = [2, 5, 3, 4, 6];
let s = "01110";
checkIfPossibleToSort(n, arr, s);
</script>
|
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
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