Find the Minimum length Unsorted Subarray, sorting which makes the complete array sorted

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:

1) 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 the indexes 3 and 8.

2) 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 the indexes 2 and 5.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Solution:
1) Find the candidate unsorted subarray
a) 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).
b) 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).

2) Check whether sorting the candidate unsorted subarray makes the complete array sorted or not. If not, then include more elements in the subarray.
a) 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.
b) 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.
c) 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)

3) Print s and e.

Implementation:

C

```#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

```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);
}
}
```

Output :
``` The unsorted subarray which makes the given array
sorted lies between the indees 3 and 8```

Time Complexity: O(n)

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.9 Average Difficulty : 2.9/5.0
Based on 169 vote(s)