Smallest subarray containing minimum and maximum values

29 Jun, 2022
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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach #include <bits/stdc++.h>using namespace std; // Function to return length of// smallest subarray containing both// maximum and minimum valueint 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 codeint main(){    int A[] = { 1, 5, 9, 7, 1, 9, 4 };    int n = sizeof(A) / sizeof(A[0]);     cout << minSubarray(A, n);     return 0;}
// Java implementation of above approachimport java.util.*; class GFG{ // Function to return length of// smallest subarray containing both// maximum and minimum valuestatic 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 codepublic 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 implementation of above approachimport sys # Function to return length of smallest# subarray containing both maximum and# minimum valuedef 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 codeA = [ 1, 5, 9, 7, 1, 9, 4 ]n = len(A) print(minSubarray(A, n)) # This code is contributed# by Saurabh_Shukla
// C# implementation of above approachusing System;using System.Linq; public class GFG{       // Function to return length of// smallest subarray containing both// maximum and minimum valuestatic 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 implementation of above approach // Function to return length of// smallest subarray containing both// maximum and minimum valuefunction 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?>
<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>
2
 
Time Complexity: O(n)Auxiliary Space: O(1)
saurabh_shukla
ankthon
vt_m
SURENDRA_GANGWAR
Dikshant Bansal
decode2207
pushpeshrajdx01
two-pointer-algorithm
Arrays
Sorting
two-pointer-algorithm
Arrays
Sorting
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