Check if given array is almost sorted (elements are at-most one position away)
Last Updated :
29 Nov, 2023
Given an array with n distinct elements. An array is said to be almost sorted (non-decreasing) if any of its elements can occur at a maximum of 1 distance away from their original places in the sorted array. We need to find whether the given array is almost sorted or not.
Examples:
Input : arr[] = {1, 3, 2, 4}
Output : Yes
Explanation : All elements are either
at original place or at most a unit away.
Input : arr[] = {1, 4, 2, 3}
Output : No
Explanation : 4 is 2 unit away from
its original place.
Sorting Approach : With the help of sorting we can predict whether our given array is almost sorted or not. The idea behind that is first sort the input array say A[]and then if array will be almost sorted then each element Ai of the given array must be equal to any of Bi-1, Bi or Bi+1 of sorted array B[].
Time Complexity : O(nlogn)
// suppose B[] is copy of A[]
sort(B, B+n);
// check first element
if ((A[0]!=B[0]) && (A[0]!=B[1]) )
return 0;
// iterate over array
for(int i=1; i<n-1; i++)
{
if (A[i]!=B[i-1]) && (A[i]!=B[i]) && (A[i]!=B[i+1]) )
return false;
}
// check for last element
if ((A[i]!=B[i-1]) && (A[i]!=B[i]) )
return 0;
// finally return true
return true;
Algorithm:
- Sort the copy of input array A to get the sorted array B.
- Check if the first element of A is either the first or second element of B. If not, return false
- Iterate through the array A from the second element to the second-to-last element
- For each element A[i] of A, check if it is equal to A[i-1], A[i], or A[i+1] in B. If not, return false
- Check if the last element of A is either the second-to-last or last element of B. If not, return false
- If all checks pass, return true.
Below is the implementation of the approach:
C++
// CPP program to find whether given array
// almost sorted or not
#include <bits/stdc++.h>
using namespace std;
// function for checking almost sort
bool isAlmostSorted(int A[], int n) {
// Make a copy of A
int B[n];
for (int i = 0; i < n; i++) {
B[i] = A[i];
}
// Sort the copy
sort(B, B+n);
// Check the first element
if ((A[0] != B[0]) && (A[0] != B[1])) {
return false;
}
// Iterate over the array
for (int i = 1; i < n-1; i++) {
if ((A[i] != B[i-1]) && (A[i] != B[i]) && (A[i] != B[i+1])) {
return false;
}
}
// Check the last element
if ((A[n-1] != B[n-1]) && (A[n-1] != B[n-2])) {
return false;
}
// Return true if the array is almost sorted
return true;
}
int main() {
// Input array
int A[] = { 1, 3, 2, 4, 6, 5 };
int n = sizeof(A)/sizeof(A[0]);
if (isAlmostSorted(A, n)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}
Java
import java.util.Arrays;
public class Main {
// Function to check if the array is almost sorted
public static boolean isAlmostSorted(int[] A, int n) {
// Make a copy of A
int[] B = Arrays.copyOf(A, n);
// Sort the copy
Arrays.sort(B);
// Check the first element
if ((A[0] != B[0]) && (A[0] != B[1])) {
return false;
}
// Iterate over the array
for (int i = 1; i < n - 1; i++) {
if ((A[i] != B[i - 1]) && (A[i] != B[i]) && (A[i] != B[i + 1])) {
return false;
}
}
// Check the last element
if ((A[n - 1] != B[n - 1]) && (A[n - 1] != B[n - 2])) {
return false;
}
// Return true if the array is almost sorted
return true;
}
public static void main(String[] args) {
// Input array
int[] A = {1, 3, 2, 4, 6, 5};
int n = A.length;
if (isAlmostSorted(A, n)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
Python3
def is_almost_sorted(arr):
# Make a copy of the array
b = arr.copy()
# Sort the copy
b.sort()
# Check the first element
if (arr[0] != b[0]) and (arr[0] != b[1]):
return False
# Iterate over the array
n = len(arr)
for i in range(1, n - 1):
if (arr[i] != b[i - 1]) and (arr[i] != b[i]) and (arr[i] != b[i + 1]):
return False
# Check the last element
if (arr[n - 1] != b[n - 1]) and (arr[n - 1] != b[n - 2]):
return False
# Return True if the array is almost sorted
return True
# Main function
if __name__ == "__main__":
# Input array
A = [1, 3, 2, 4, 6, 5]
if is_almost_sorted(A):
print("Yes")
else:
print("No")
C#
using System;
using System.Linq;
class Program {
// Function for checking almost sorted
static bool IsAlmostSorted(int[] A, int n)
{
// Make a copy of A
int[] B = new int[n];
Array.Copy(A, B, n);
// Sort the copy
Array.Sort(B);
// Check the first element
if ((A[0] != B[0]) && (A[0] != B[1])) {
return false;
}
// Iterate over the array
for (int i = 1; i < n - 1; i++) {
if ((A[i] != B[i - 1]) && (A[i] != B[i])
&& (A[i] != B[i + 1])) {
return false;
}
}
// Check the last element
if ((A[n - 1] != B[n - 1])
&& (A[n - 1] != B[n - 2])) {
return false;
}
// Return true if the array is almost sorted
return true;
}
static void Main()
{
// Input array
int[] A = { 1, 3, 2, 4, 6, 5 };
int n = A.Length;
if (IsAlmostSorted(A, n)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
JavaScript
// function for checking almost sort
function isAlmostSorted(A, n) {
// Make a copy of A
let B = [...A];
// Sort the copy
B.sort((a, b) => a - b);
// Check the first element
if ((A[0] != B[0]) && (A[0] != B[1])) {
return false;
}
// Iterate over the array
for (let i = 1; i < n-1; i++) {
if ((A[i] != B[i-1]) && (A[i] != B[i]) && (A[i] != B[i+1])) {
return false;
}
}
// Check the last element
if ((A[n-1] != B[n-1]) && (A[n-1] != B[n-2])) {
return false;
}
// Return true if the array is almost sorted
return true;
}
// Input array
let A = [1, 3, 2, 4, 6, 5];
let n = A.length;
if (isAlmostSorted(A, n)) {
console.log("Yes");
} else {
console.log("No");
}
Time complexity: O(n Log n)
Auxiliary Space: O(n) as an array B has been created to store copy of input array A. Here, n is size of the input array.
Efficient Approach: The idea is based on Bubble Sort. Like Bubble Sort, we compare adjacent elements and swap them if they are not in order. Here after swapping we move the index one position extra so that bubbling is limited to one place. So after one iteration if the resultant array is sorted then we can say that our input array was almost sorted otherwise not almost sorted.
// perform bubble sort tech once
for (int i=0; i<n-1; i++)
if (A[i+1]<A[i])
swap(A[i], A[i+1]);
i++;
// check whether resultant is sorted or not
for (int i=0; i<n-1; i++)
if (A[i+1]<A[i])
return false;
// If resultant is sorted return true
return true;
C++
// CPP program to find whether given array
// almost sorted or not
#include <bits/stdc++.h>
using namespace std;
// function for checking almost sort
bool almostSort(int A[], int n)
{
// One by one compare adjacents.
for (int i = 0; i < n - 1; i++) {
if (A[i] > A[i + 1]) {
swap(A[i], A[i + 1]);
i++;
}
}
// check whether resultant is sorted or not
for (int i = 0; i < n - 1; i++)
if (A[i] > A[i + 1])
return false;
// is resultant is sorted return true
return true;
}
// driver function
int main()
{
int A[] = { 1, 3, 2, 4, 6, 5 };
int n = sizeof(A) / sizeof(A[0]);
if (almostSort(A, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// JAVA Code to check if given array is almost
// sorted or not
import java.util.*;
class GFG {
// function for checking almost sort
public static boolean almostSort(int A[], int n)
{
// One by one compare adjacents.
for (int i = 0; i < n - 1; i++) {
if (A[i] > A[i + 1]) {
int temp = A[i];
A[i] = A[i+1];
A[i+1] = temp;
i++;
}
}
// check whether resultant is sorted or not
for (int i = 0; i < n - 1; i++)
if (A[i] > A[i + 1])
return false;
// is resultant is sorted return true
return true;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int A[] = { 1, 3, 2, 4, 6, 5 };
int n = A.length;
if (almostSort(A, n))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find whether given
# array almost sorted or not
# Function for checking almost sort
def almostSort(A, n):
# One by one compare adjacents.
i = 0
while i < n - 1:
if A[i] > A[i + 1]:
A[i], A[i + 1] = A[i + 1], A[i]
i += 1
i += 1
# check whether resultant is sorted or not
for i in range(0, n - 1):
if A[i] > A[i + 1]:
return False
# Is resultant is sorted return true
return True
# Driver Code
if __name__ == "__main__":
A = [1, 3, 2, 4, 6, 5]
n = len(A)
if almostSort(A, n):
print("Yes")
else:
print("No")
# This code is contributed
# by Rituraj Jain
C#
// C# Code to check if given array
// is almost sorted or not
using System;
class GFG {
// function for checking almost sort
public static bool almostSort(int []A, int n)
{
// One by one compare adjacents.
for (int i = 0; i < n - 1; i++)
{
if (A[i] > A[i + 1])
{
int temp = A[i];
A[i] = A[i + 1];
A[i + 1] = temp;
i++;
}
}
// Check whether resultant is
// sorted or not
for (int i = 0; i < n - 1; i++)
if (A[i] > A[i + 1])
return false;
// is resultant is sorted return true
return true;
}
// Driver Code
public static void Main()
{
int []A = {1, 3, 2, 4, 6, 5};
int n = A.Length;
if (almostSort(A, n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Nitin Mittal.
JavaScript
<script>
// Javascript Code to check if given array
// is almost sorted or not
// function for checking almost sort
function almostSort(A, n)
{
// One by one compare adjacents.
for (let i = 0; i < n - 1; i++)
{
if (A[i] > A[i + 1])
{
let temp = A[i];
A[i] = A[i + 1];
A[i + 1] = temp;
i++;
}
}
// Check whether resultant is
// sorted or not
for (let i = 0; i < n - 1; i++)
if (A[i] > A[i + 1])
return false;
// is resultant is sorted return true
return true;
}
let A = [1, 3, 2, 4, 6, 5];
let n = A.length;
if (almostSort(A, n))
document.write("Yes");
else
document.write("No");
</script>
PHP
<?php
// PHP program to find
// whether given array
// almost sorted or not
// function for checking
// almost sort
function almostSort($A, $n)
{
// One by one compare adjacents.
for ($i = 0; $i < $n - 1; $i++)
{
if ($A[$i] > $A[$i + 1])
{
list($A[$i],
$A[$i + 1]) = array($A[$i + 1],
$A[$i] );
$i++;
}
}
// check whether resultant
// is sorted or not
for ($i = 0; $i <$n - 1; $i++)
if ($A[$i] > $A[$i + 1])
return false;
// is resultant is
// sorted return true
return true;
}
// Driver Code
$A = array (1, 3, 2,
4, 6, 5);
$n = sizeof($A) ;
if (almostSort($A, $n))
echo "Yes", "\n";
else
echo "Yes", "\n";
// This code is contributed by ajit
?>
Output:
Yes
Time Complexity: O(N), as we are using any loops for traversing.
Auxiliary Space: O(1), as we are not using any extra space.
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