Given an unsorted array arr[0..n-1] of size n, find the minimum length subarray arr[s..e] such that sorting this subarray makes the whole array sorted.
Examples:
- If the input array is [10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60], your program should be able to find that the subarray lies between indexes 3 and 8.
- If the input array is [0, 1, 15, 25, 6, 7, 30, 40, 50], your program should be able to find that the subarray lies between indexes 2 and 5.
Approach 1:
Idea/Intuition :
Make a temporary array same as the given array ,sort the temporary array . Now check from starting at which index the element of the given array and temporary array are unequal and store it in temporary variable s . Repeat the above From the end and store the index at another temporary variable e . The length e-s+1 is the length of smallest unequal subarray .
Algorithm :
- Declare a temporary array temp same as given array arr.
- Sort the temporary array .
- Initialize variable s with 0 and e with 0.
- Checking the unequal element from start and storing it in s variable .
- Checking the equal element from end and storing it in e variable.
- Returning (e-s+1) .
- Printing the result .
Below is the implementation of above approach .
Code :
C++
#include <bits/stdc++.h> using namespace std; // function performing calculation int minLength(vector< int >& arr) { // temporary array equal to given array vector< int > temp = arr; // sorting the temporary array sort(temp.begin(), temp.end()); // initializing indices int s = 0, e = 0; // checking the unequal element from start and storing // it in s variable for ( int i = 0; i < arr.size(); i++) { if (arr[i] != temp[i]) { s = i; break ; } } // checking the unequal element from end and storing it // in e variable for ( int i = arr.size() - 1; i >= 0; i--) { if (arr[i] != temp[i]) { e = i; break ; } } // returning minimum length return (e - s + 1); } // driver function int main() { // given array arr vector< int > arr = { 10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60 }; // calling the function performing calculation and // printing the result cout << "Minimum length of subarray is : " << minLength(arr); return 0; } |
Python3
# importing the library from typing import List # function performing calculation def minLength(arr: List [ int ]) - > int : # temporary array equal to given array temp = arr[:] # sorting the temporary array temp.sort() # initializing indices s = 0 e = 0 # checking the unequal element from start and storing # it in s variable for i in range ( len (arr)): if arr[i] ! = temp[i]: s = i break # checking the unequal element from end and storing it # in e variable for i in range ( len (arr) - 1 , - 1 , - 1 ): if arr[i] ! = temp[i]: e = i break # returning minimum length return (e - s + 1 ) # driver function if __name__ = = '__main__' : # given array arr arr = [ 10 , 12 , 20 , 30 , 25 , 40 , 32 , 31 , 35 , 50 , 60 ] # calling the function performing calculation and # printing the result print ( "Minimum length of subarray is : " , minLength(arr)) # This code is contributed by divyansh2212 |
C#
using System; using System.Linq; class Program { //function performing calculations static int MinLength( int [] arr) { //temporary array equal to given array int [] temp = arr.ToArray(); //sorting the temporary array Array.Sort(temp); //initializing indices int s = 0; int e = 0; //checking the unequal element from start and storing it in s variable for ( int i = 0; i < arr.Length; i++) { if (arr[i] != temp[i]) { s = i; break ; } } //checking the unequal elements from end and storing it in e variable for ( int i = arr.Length - 1; i >= 0; i--) { if (arr[i] != temp[i]) { e = i; break ; } } //returning minimum length return e - s + 1; } static void Main( string [] args) { int [] arr = new int [] { 10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60 }; Console.WriteLine( "Minimum length of subarray is : " + MinLength(arr)); } } //This code is contributed by snehalsalokhe |
Java
import java.util.*; public class GFG { // function performing calculation public static int minLength(ArrayList<Integer> arr) { // temporary array equal to given array ArrayList<Integer> temp = new ArrayList<Integer>(arr); // sorting the temporary array Collections.sort(temp); // initializing indices int s = 0 , e = 0 ; // checking the unequal element from start and storing // it in s variable for ( int i = 0 ; i < arr.size(); i++) { if (arr.get(i) != temp.get(i)) { s = i; break ; } } // checking the unequal element from end and storing it // in e variable for ( int i = arr.size() - 1 ; i >= 0 ; i--) { if (arr.get(i) != temp.get(i)) { e = i; break ; } } // returning minimum length return (e - s + 1 ); } // driver function public static void main(String[] args) { // given array arr ArrayList<Integer> arr = new ArrayList<Integer>( Arrays.asList( 10 , 12 , 20 , 30 , 25 , 40 , 32 , 31 , 35 , 50 , 60 ) ); // calling the function performing calculation and // printing the result System.out.println( "Minimum length of subarray is : " + minLength(arr)); } } |
Javascript
// function performing calculation function minLength(arr) { // temporary array equal to given array let temp = []; for (let i=0;i<arr.length;i++) { temp.push(arr[i]); } // sorting the temporary array temp.sort(); // initializing indices let s = 0, e = 0; // checking the unequal element from start and storing // it in s variable for (let i = 0; i < arr.length; i++) { if (arr[i] != temp[i]) { s = i; break ; } } // checking the unequal element from end and storing it // in e variable for (let i = arr.length - 1; i >= 0; i--) { if (arr[i] != temp[i]) { e = i; break ; } } // returning minimum length return (e - s + 1); } // driver function // given array arr let arr = [ 10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60 ]; // calling the function performing calculation and // printing the result console.log( "Minimum length of subarray is : " + minLength(arr)); // This code is contributed by akashish__ |
Output
Minimum length of subarray is : 6
Time Complexity : O(NLog(N)) , where N is the size of given array
Space Complexity : O(N) , Space for temporary array temp .
Approach 2:
- Find the candidate unsorted subarray
- Scan from left to right and find the first element which is greater than the next element. Let s be the index of such an element. In the above example 1, s is 3 (index of 30).
- Scan from right to left and find the first element (first in right to left order) which is smaller than the next element (next in right to left order). Let e be the index of such an element. In the above example 1, e is 7 (index of 31).
- Check whether sorting the candidate unsorted subarray makes the complete array sorted or not. If not, then include more elements in the subarray.
- Find the minimum and maximum values in arr[s..e]. Let minimum and maximum values be min and max. min and max for [30, 25, 40, 32, 31] are 25 and 40 respectively.
- Find the first element (if there is any) in arr[0..s-1] which is greater than min, change s to index of this element. There is no such element in above example 1.
- Find the last element (if there is any) in arr[e+1..n-1] which is smaller than max, change e to index of this element. In the above example 1, e is changed to 8 (index of 35)
- Print s and e.
Below is the implementation of the above approach:
C++
// C++ program to find the Minimum length Unsorted Subarray, // sorting which makes the complete array sorted #include<bits/stdc++.h> using namespace std; void printUnsorted( int arr[], int n) { int s = 0, e = n-1, i, max, min; // step 1(a) of above algo for (s = 0; s < n-1; s++) { if (arr[s] > arr[s+1]) break ; } if (s == n-1) { cout << "The complete array is sorted" ; return ; } // step 1(b) of above algo for (e = n - 1; e > 0; e--) { if (arr[e] < arr[e-1]) break ; } // step 2(a) of above algo max = arr[s]; min = arr[s]; for (i = s + 1; i <= e; i++) { if (arr[i] > max) max = arr[i]; if (arr[i] < min) min = arr[i]; } // step 2(b) of above algo for ( i = 0; i < s; i++) { if (arr[i] > min) { s = i; break ; } } // step 2(c) of above algo for ( i = n -1; i >= e+1; i--) { if (arr[i] < max) { e = i; break ; } } // step 3 of above algo cout << "The unsorted subarray which" << " makes the given array" << endl << "sorted lies between the indices " << s << " and " << e; return ; } int main() { int arr[] = {10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60}; int arr_size = sizeof (arr)/ sizeof (arr[0]); printUnsorted(arr, arr_size); getchar (); return 0; } // This code is contributed // by Akanksha Rai |
C
// C program to find the Minimum length Unsorted Subarray, // sorting which makes the complete array sorted #include<stdio.h> void printUnsorted( int arr[], int n) { int s = 0, e = n-1, i, max, min; // step 1(a) of above algo for (s = 0; s < n-1; s++) { if (arr[s] > arr[s+1]) break ; } if (s == n-1) { printf ( "The complete array is sorted" ); return ; } // step 1(b) of above algo for (e = n - 1; e > 0; e--) { if (arr[e] < arr[e-1]) break ; } // step 2(a) of above algo max = arr[s]; min = arr[s]; for (i = s + 1; i <= e; i++) { if (arr[i] > max) max = arr[i]; if (arr[i] < min) min = arr[i]; } // step 2(b) of above algo for ( i = 0; i < s; i++) { if (arr[i] > min) { s = i; break ; } } // step 2(c) of above algo for ( i = n -1; i >= e+1; i--) { if (arr[i] < max) { e = i; break ; } } // step 3 of above algo printf ( " The unsorted subarray which makes the given array " " sorted lies between the indees %d and %d" , s, e); return ; } int main() { int arr[] = {10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60}; int arr_size = sizeof (arr)/ sizeof (arr[0]); printUnsorted(arr, arr_size); getchar (); return 0; } |
Java
// Java program to find the Minimum length Unsorted Subarray, // sorting which makes the complete array sorted import java.io.*; class Main { static void printUnsorted( int arr[], int n) { int s = 0 , e = n- 1 , i, max, min; // step 1(a) of above algo for (s = 0 ; s < n- 1 ; s++) { if (arr[s] > arr[s+ 1 ]) break ; } if (s == n- 1 ) { System.out.println( "The complete array is sorted" ); return ; } // step 1(b) of above algo for (e = n - 1 ; e > 0 ; e--) { if (arr[e] < arr[e- 1 ]) break ; } // step 2(a) of above algo max = arr[s]; min = arr[s]; for (i = s + 1 ; i <= e; i++) { if (arr[i] > max) max = arr[i]; if (arr[i] < min) min = arr[i]; } // step 2(b) of above algo for ( i = 0 ; i < s; i++) { if (arr[i] > min) { s = i; break ; } } // step 2(c) of above algo for ( i = n - 1 ; i >= e+ 1 ; i--) { if (arr[i] < max) { e = i; break ; } } // step 3 of above algo System.out.println( " The unsorted subarray which" + " makes the given array sorted lies" + " between the indices " +s+ " and " +e); return ; } public static void main(String args[]) { int arr[] = { 10 , 12 , 20 , 30 , 25 , 40 , 32 , 31 , 35 , 50 , 60 }; int arr_size = arr.length; printUnsorted(arr, arr_size); } } |
Python3
# Python3 program to find the Minimum length Unsorted Subarray, # sorting which makes the complete array sorted def printUnsorted(arr, n): e = n - 1 # step 1(a) of above algo for s in range ( 0 ,n - 1 ): if arr[s] > arr[s + 1 ]: break if s = = n - 1 : print ( "The complete array is sorted" ) exit() # step 1(b) of above algo e = n - 1 while e > 0 : if arr[e] < arr[e - 1 ]: break e - = 1 # step 2(a) of above algo max = arr[s] min = arr[s] for i in range (s + 1 ,e + 1 ): if arr[i] > max : max = arr[i] if arr[i] < min : min = arr[i] # step 2(b) of above algo for i in range (s): if arr[i] > min : s = i break # step 2(c) of above algo i = n - 1 while i > = e + 1 : if arr[i] < max : e = i break i - = 1 # step 3 of above algo print ( "The unsorted subarray which makes the given array" ) print ( "sorted lies between the indexes %d and %d" % ( s, e)) arr = [ 10 , 12 , 20 , 30 , 25 , 40 , 32 , 31 , 35 , 50 , 60 ] arr_size = len (arr) printUnsorted(arr, arr_size) # This code is contributed by Shreyanshi Arun |
C#
// C# program to find the Minimum length Unsorted Subarray, // sorting which makes the complete array sorted using System; class GFG { static void printUnsorted( int []arr, int n) { int s = 0, e = n-1, i, max, min; // step 1(a) of above algo for (s = 0; s < n-1; s++) { if (arr[s] > arr[s+1]) break ; } if (s == n-1) { Console.Write( "The complete " + "array is sorted" ); return ; } // step 1(b) of above algo for (e = n - 1; e > 0; e--) { if (arr[e] < arr[e-1]) break ; } // step 2(a) of above algo max = arr[s]; min = arr[s]; for (i = s + 1; i <= e; i++) { if (arr[i] > max) max = arr[i]; if (arr[i] < min) min = arr[i]; } // step 2(b) of above algo for ( i = 0; i < s; i++) { if (arr[i] > min) { s = i; break ; } } // step 2(c) of above algo for ( i = n -1; i >= e+1; i--) { if (arr[i] < max) { e = i; break ; } } // step 3 of above algo Console.Write( " The unsorted subarray which" + " makes the given array sorted lies \n" + " between the indices " +s+ " and " +e); return ; } public static void Main() { int []arr = {10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60}; int arr_size = arr.Length; printUnsorted(arr, arr_size); } } // This code contributed by Sam007 |
PHP
<?php // PHP program to find the Minimum length Unsorted Subarray, // sorting which makes the complete array sorted function printUnsorted(& $arr , $n ) { $s = 0; $e = $n - 1; // step 1(a) of above algo for ( $s = 0; $s < $n - 1; $s ++) { if ( $arr [ $s ] > $arr [ $s + 1]) break ; } if ( $s == $n - 1) { echo "The complete array is sorted" ; return ; } // step 1(b) of above algo for ( $e = $n - 1; $e > 0; $e --) { if ( $arr [ $e ] < $arr [ $e - 1]) break ; } // step 2(a) of above algo $max = $arr [ $s ]; $min = $arr [ $s ]; for ( $i = $s + 1; $i <= $e ; $i ++) { if ( $arr [ $i ] > $max ) $max = $arr [ $i ]; if ( $arr [ $i ] < $min ) $min = $arr [ $i ]; } // step 2(b) of above algo for ( $i = 0; $i < $s ; $i ++) { if ( $arr [ $i ] > $min ) { $s = $i ; break ; } } // step 2(c) of above algo for ( $i = $n - 1; $i >= $e + 1; $i --) { if ( $arr [ $i ] < $max ) { $e = $i ; break ; } } // step 3 of above algo echo " The unsorted subarray which makes " . "the given array " . "\n" . " sorted lies between the indees " . $s . " and " . $e ; return ; } // Driver code $arr = array (10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60); $arr_size = sizeof( $arr ); printUnsorted( $arr , $arr_size ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to find the Minimum length Unsorted Subarray, // sorting which makes the complete array sorted function printUnsorted(arr,n) { let s = 0, e = n-1, i, max, min; // step 1(a) of above algo for (s = 0; s < n-1; s++) { if (arr[s] > arr[s+1]) break ; } if (s == n-1) { document.write( "The complete array is sorted" ); return ; } // step 1(b) of above algo for (e = n - 1; e > 0; e--) { if (arr[e] < arr[e-1]) break ; } // step 2(a) of above algo max = arr[s]; min = arr[s]; for (i = s + 1; i <= e; i++) { if (arr[i] > max) max = arr[i]; if (arr[i] < min) min = arr[i]; } // step 2(b) of above algo for ( i = 0; i < s; i++) { if (arr[i] > min) { s = i; break ; } } // step 2(c) of above algo for ( i = n -1; i >= e+1; i--) { if (arr[i] < max) { e = i; break ; } } // step 3 of above algo document.write( " The unsorted subarray which" + " makes the given array sorted lies" + " between the indees " +s+ " and " +e); return ; } let arr=[10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60]; let arr_size = arr.length; printUnsorted(arr, arr_size); // This code is contributed by avanitrachhadiya2155 </script> |
Output
The unsorted subarray which makes the given array sorted lies between the indices 3 and 8
Time Complexity : O(n)
Auxiliary Space : O(1)
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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