Smallest subarray containing minimum and maximum values

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Given an array A of size N. The task is to find the length of smallest subarray which contains both maximum and minimum values.

Examples:

Input : A[] = {1, 5, 9, 7, 1, 9, 4}
Output : 2
subarray {1, 9} has both maximum and minimum value.

Input : A[] = {2, 2, 2, 2}
Output : 1
2 is both maximum and minimum here.

Approach: The idea is to use two-pointer technique here :

Find the maximum and minimum values of the array.
Traverse through the array and store the last occurrences of maximum and minimum values.
If the of last occurrence of maximum is pos_max and minimum is pos_min, then the minimum value of abs(pos_min – pos_max) + 1 is our required answer.

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return length of
// smallest subarray containing both
// maximum and minimum value
int minSubarray(int A[], int n)
{
 
    // find maximum and minimum
    // values in the array
    int minValue = *min_element(A, A + n);
    int maxValue = *max_element(A, A + n);
 
    int pos_min = -1, pos_max = -1, ans = INT_MAX;
 
    // iterate over the array and set answer
    // to smallest difference between position
    // of maximum and position of minimum value
    for (int i = 0; i < n; i++) {
 
        // last occurrence of minValue
        if (A[i] == minValue)
            pos_min = i;
 
        // last occurrence of maxValue
        if (A[i] == maxValue)
            pos_max = i;
 
        if (pos_max != -1 and pos_min != -1)
            ans = min(ans, abs(pos_min - pos_max) + 1);
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int A[] = { 1, 5, 9, 7, 1, 9, 4 };
    int n = sizeof(A) / sizeof(A[0]);
 
    cout << minSubarray(A, n);
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG
{
 
// Function to return length of
// smallest subarray containing both
// maximum and minimum value
static int minSubarray(int A[], int n)
{
 
    // find maximum and minimum
    // values in the array
    int minValue = A[0];
    for(int i = 1; i < n; i++)
    {
        if(A[i] < minValue)
            minValue = A[i];
    }
    int maxValue = A[0];
    for(int i = 1; i < n; i++)
    {
        if(A[i] > maxValue)
            maxValue = A[i];
    }
 
    int pos_min = -1, pos_max = -1, 
        ans = Integer.MAX_VALUE;
 
    // iterate over the array and set answer
    // to smallest difference between position
    // of maximum and position of minimum value
    for (int i = 0; i < n; i++) 
    {
 
        // last occurrence of minValue
        if (A[i] == minValue)
            pos_min = i;
 
        // last occurrence of maxValue
        if (A[i] == maxValue)
            pos_max = i;
 
        if (pos_max != -1 && pos_min != -1)
            ans = Math.min(ans, 
                  Math.abs(pos_min - pos_max) + 1);
    }
 
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    int A[] = { 1, 5, 9, 7, 1, 9, 4 };
    int n = A.length;
 
    System.out.println(minSubarray(A, n));
}
}
 
// This code is contributed by
// Surendra_Gangwar

Python3

# Python3 implementation of above approach
import sys
 
# Function to return length of smallest 
# subarray containing both maximum and 
# minimum value
def minSubarray(A, n):
 
    # find maximum and minimum
    # values in the array
    minValue = min(A)
    maxValue = max(A)
 
    pos_min, pos_max, ans = -1, -1, sys.maxsize
 
    # iterate over the array and set answer
    # to smallest difference between position
    # of maximum and position of minimum value
    for i in range(0, n):
         
        # last occurrence of minValue
        if A[i] == minValue:
            pos_min = i
 
        # last occurrence of maxValue
        if A[i] == maxValue:
            pos_max = i
 
        if pos_max != -1 and pos_min != -1 :
            ans = min(ans, abs(pos_min - pos_max) + 1)
 
    return ans
 
# Driver code
A = [ 1, 5, 9, 7, 1, 9, 4 ]
n = len(A)
 
print(minSubarray(A, n))
 
# This code is contributed
# by Saurabh_Shukla

C#

// C# implementation of above approach
using System;
using System.Linq;
 
public class GFG{
     
 
 
// Function to return length of
// smallest subarray containing both
// maximum and minimum value
static int minSubarray(int []A, int n)
{
 
    // find maximum and minimum
    // values in the array
    int minValue = A.Min();
    int maxValue = A.Max();
 
    int pos_min = -1, pos_max = -1, ans = int.MaxValue;
 
    // iterate over the array and set answer
    // to smallest difference between position
    // of maximum and position of minimum value
    for (int i = 0; i < n; i++) {
 
        // last occurrence of minValue
        if (A[i] == minValue)
            pos_min = i;
 
        // last occurrence of maxValue
        if (A[i] == maxValue)
            pos_max = i;
 
        if (pos_max != -1 && pos_min != -1)
            ans = Math.Min(ans, Math.Abs(pos_min - pos_max) + 1);
    }
 
    return ans;
}
 
// Driver code
 
 
    static public void Main (){
            int []A = { 1, 5, 9, 7, 1, 9, 4 };
    int n = A.Length;
 
    Console.WriteLine(minSubarray(A, n));
    }
}
// This code is contributed by anuj_67..

PHP

<?php
// PHP implementation of above approach 
 
// Function to return length of 
// smallest subarray containing both 
// maximum and minimum value 
function minSubarray($A, $n) 
{ 
 
    // find maximum and minimum 
    // values in the array 
    $minValue = min($A);
    $maxValue = max($A);
 
    $pos_min = -1;
    $pos_max = -1;
    $ans = PHP_INT_MAX;
 
    // iterate over the array and set answer 
    // to smallest difference between position 
    // of maximum and position of minimum value 
    for ($i = 0; $i < $n; $i++)
    { 
 
        // last occurrence of minValue 
        if ($A[$i] == $minValue) 
            $pos_min = $i; 
 
        // last occurrence of maxValue 
        if ($A[$i] == $maxValue) 
            $pos_max = $i; 
 
        if ($pos_max != -1 and $pos_min != -1) 
            $ans = min($ans, abs($pos_min - 
                                 $pos_max) + 1); 
    } 
 
    return $ans; 
} 
 
// Driver code 
$A = array(1, 5, 9, 7, 1, 9, 4);
$n = sizeof($A);
 
echo minSubarray($A, $n); 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
    // Javascript implementation of above approach
     
    // Function to return length of
    // smallest subarray containing both
    // maximum and minimum value
    function minSubarray(A, n)
    {
 
        // find maximum and minimum
        // values in the array
        let minValue = Number.MAX_VALUE;
        let maxValue = Number.MIN_VALUE;
        for (let i = 0; i < n; i++) 
        {
            minValue = Math.min(minValue, A[i]);
            maxValue = Math.max(maxValue, A[i]);
        }
 
        let pos_min = -1, pos_max = -1, ans = Number.MAX_VALUE;
 
        // iterate over the array and set answer
        // to smallest difference between position
        // of maximum and position of minimum value
        for (let i = 0; i < n; i++) {
 
            // last occurrence of minValue
            if (A[i] == minValue)
                pos_min = i;
 
            // last occurrence of maxValue
            if (A[i] == maxValue)
                pos_max = i;
 
            if (pos_max != -1 && pos_min != -1)
                ans = Math.min(ans, Math.abs(pos_min - pos_max) + 1);
        }
 
        return ans;
    }
     
    let A = [ 1, 5, 9, 7, 1, 9, 4 ];
    let n = A.length;
   
    document.write(minSubarray(A, n));
 
</script>

Output

Complexity Analysis:

Time Complexity: O(n)
Auxiliary Space: O(1)

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