Min difference between maximum and minimum element in all Y size subarrays

Difficulty Level : Medium
Last Updated : 10 Jun, 2021

Given an array arr[] of size N and integer Y, the task is to find a minimum of all the differences between the maximum and minimum elements in all the sub-arrays of size Y.

Examples:

Input: arr[] = { 3, 2, 4, 5, 6, 1, 9 } Y = 3
Output: 2
Explanation:
All subarrays of length = 3 are:
{3, 2, 4} where maximum element = 4 and minimum element = 2 difference = 2
{2, 4, 5} where maximum element = 5 and minimum element = 2 difference = 3
{4, 5, 6} where maximum element = 6 and minimum element = 4 difference = 2
{5, 6, 1} where maximum element = 6 and minimum element = 1 difference = 5
{6, 1, 9} where maximum element = 9 and minimum element = 6 difference = 3
Out of these, the minimum is 2.

Input: arr[] = { 1, 2, 3, 3, 2, 2 } Y = 4
Output: 1
Explanation:
All subarrays of length = 4 are:
{1, 2, 3, 3} maximum element = 3 and minimum element = 1 difference = 2
{2, 3, 3, 2} maximum element = 3 and minimum element = 2 difference = 1
{3, 3, 2, 2} maximum element = 3 and minimum element = 2 difference = 1
Out of these, the minimum is 1.

Naive Approach: The naive idea is to traverse for every index i in the range [0, N – Y] use another loop to traverse from i^th index to (i + Y – 1)^thindex and then calculate the minimum and maximum elements in a subarray of size Y and hence calculate the difference of maximum and minimum element for that i^th iteration. Finally, by keeping a check on differences, evaluate the minimum difference.

Time Complexity: O(N*Y)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use the concept of the approach discussed in the Next Greater Element article. Below are the steps:

Build two arrays maxarr[] and minarr[], where maxarr[] will store the index of the element which is next greater to the element at i^th index and minarr[] will store the index of the next element which is less than the element at the i^th index.
Initialize a stack with 0 to store the indices in both the above cases.
For each index, i in the range [0, N – Y], using a nested loop and sliding window approach form two arrays submax and submin. These arrays will store maximum and minimum elements in the subarray in i^th iteration.
Finally, calculate the minimum difference.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to get the maximum of all
// the subarrays of size Y
vector<int> get_submaxarr(int* arr,
                        int n, int y)
{
    int j = 0;
    stack<int> stk;
 
    // ith index of maxarr array
    // will be the index upto which
    // Arr[i] is maximum
    vector<int> maxarr(n);
    stk.push(0);
 
    for (int i = 1; i < n; i++) {
 
        // Stack is used to find the
        // next larger element and
        // keeps track of index of
        // current iteration
        while (stk.empty() == false
            and arr[i] > arr[stk.top()]) {
 
            maxarr[stk.top()] = i - 1;
            stk.pop();
        }
        stk.push(i);
    }
 
    // Loop for remaining indexes
    while (!stk.empty()) {
 
        maxarr[stk.top()] = n - 1;
        stk.pop();
    }
    vector<int> submax;
 
    for (int i = 0; i <= n - y; i++) {
 
        // j < i used to keep track
        // whether jth element is
        // inside or outside the window
        while (maxarr[j] < i + y - 1
            or j < i) {
            j++;
        }
 
        submax.push_back(arr[j]);
    }
 
    // Return submax
    return submax;
}
 
// Function to get the minimum for
// all subarrays of size Y
vector<int> get_subminarr(int* arr,
                        int n, int y)
{
    int j = 0;
 
    stack<int> stk;
 
    // ith index of minarr array
    // will be the index upto which
    // Arr[i] is minimum
    vector<int> minarr(n);
    stk.push(0);
 
    for (int i = 1; i < n; i++) {
 
        // Stack is used to find the
        // next smaller element and
        // keeping track of index of
        // current iteration
        while (stk.empty() == false
            and arr[i] < arr[stk.top()]) {
 
            minarr[stk.top()] = i;
            stk.pop();
        }
        stk.push(i);
    }
 
    // Loop for remaining indexes
    while (!stk.empty()) {
 
        minarr[stk.top()] = n;
        stk.pop();
    }
 
    vector<int> submin;
 
    for (int i = 0; i <= n - y; i++) {
 
        // j < i used to keep track
        // whether jth element is inside
        // or outside the window
 
        while (minarr[j] <= i + y - 1
            or j < i) {
            j++;
        }
 
        submin.push_back(arr[j]);
    }
 
    // Return submin
    return submin;
}
 
// Function to get minimum difference
void getMinDifference(int Arr[], int N,
                    int Y)
{
    // Create submin and submax arrays
    vector<int> submin
        = get_subminarr(Arr, N, Y);
 
    vector<int> submax
        = get_submaxarr(Arr, N, Y);
 
    // Store initial difference
    int minn = submax[0] - submin[0];
    int b = submax.size();
 
    for (int i = 1; i < b; i++) {
 
        // Calculate temporary difference
        int dif = submax[i] - submin[i];
        minn = min(minn, dif);
    }
 
    // Final minimum difference
    cout << minn << "\n";
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 3, 2, 2 };
    int N = sizeof arr / sizeof arr[0];
 
    // Given subarray size
    int Y = 4;
 
    // Function Call
    getMinDifference(arr, N, Y);
    return 0;
}

Python3

# Python3 program for the above approach
 
# Function to get the maximum of all
# the subarrays of size Y
def get_submaxarr(arr, n, y):
     
    j = 0
    stk = []
     
    # ith index of maxarr array
    # will be the index upto which
    # Arr[i] is maximum
    maxarr = [0] * n
    stk.append(0)
 
    for i in range(1, n):
 
        # Stack is used to find the
        # next larger element and
        # keeps track of index of
        # current iteration
        while (len(stk) > 0 and
                 arr[i] > arr[stk[-1]]):
            maxarr[stk[-1]] = i - 1
            stk.pop()
             
        stk.append(i)
 
    # Loop for remaining indexes
    while (stk):
        maxarr[stk[-1]] = n - 1
        stk.pop()
         
    submax = []
     
    for i in range(n - y + 1):
 
        # j < i used to keep track
        # whether jth element is
        # inside or outside the window
        while (maxarr[j] < i + y - 1 or
                       j < i):
            j += 1
             
        submax.append(arr[j])
 
    # Return submax
    return submax
 
# Function to get the minimum for
# all subarrays of size Y
def get_subminarr(arr, n, y):
     
    j = 0
    stk = []
     
    # ith index of minarr array
    # will be the index upto which
    # Arr[i] is minimum
    minarr = [0] * n
    stk.append(0)
     
    for i in range(1 , n):
         
        # Stack is used to find the
        # next smaller element and
        # keeping track of index of
        # current iteration
        while (stk and arr[i] < arr[stk[-1]]):
            minarr[stk[-1]] = i
            stk.pop()
             
        stk.append(i)
         
    # Loop for remaining indexes
    while (stk):
        minarr[stk[-1]] = n
        stk.pop()
         
    submin = []
     
    for i in range(n - y + 1):
         
        # j < i used to keep track
        # whether jth element is inside
        # or outside the window
        while (minarr[j] <= i + y - 1 or
                      j < i):
            j += 1
             
        submin.append(arr[j])
         
    # Return submin
    return submin
 
# Function to get minimum difference
def getMinDifference(Arr, N, Y):
     
    # Create submin and submax arrays
    submin = get_subminarr(Arr, N, Y)
    submax = get_submaxarr(Arr, N, Y)
     
    # Store initial difference
    minn = submax[0] - submin[0]
    b = len(submax)
     
    for i in range(1, b):
         
        # Calculate temporary difference
        diff = submax[i] - submin[i]
        minn = min(minn, diff)
     
    # Final minimum difference
    print(minn)
     
# Driver code
 
# Given array arr[]
arr = [ 1, 2, 3, 3, 2, 2 ]
N = len(arr)
 
# Given subarray size
Y = 4
 
# Function call
getMinDifference(arr, N, Y)
     
# This code is contributed by Stuti Pathak

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to get the maximum of all
// the subarrays of size Y
function get_submaxarr(arr, n, y)
{
    var j = 0;
    var stk = [];
 
    // ith index of maxarr array
    // will be the index upto which
    // Arr[i] is maximum
    var maxarr = Array(n);
    stk.push(0);
 
    for (var i = 1; i < n; i++) {
 
        // Stack is used to find the
        // next larger element and
        // keeps track of index of
        // current iteration
        while (stk.length!=0
            && arr[i] > arr[stk[stk.length-1]]) {
 
            maxarr[stk[stk.length-1]] = i - 1;
            stk.pop();
        }
        stk.push(i);
    }
 
    // Loop for remaining indexes
    while (stk.length!=0) {
 
        maxarr[stk[stk.length-1]] = n - 1;
        stk.pop();
    }
    var submax = [];
 
    for (var i = 0; i <= n - y; i++) {
 
        // j < i used to keep track
        // whether jth element is
        // inside or outside the window
        while (maxarr[j] < i + y - 1
            || j < i) {
            j++;
        }
 
        submax.push(arr[j]);
    }
 
    // Return submax
    return submax;
}
 
// Function to get the minimum for
// all subarrays of size Y
function get_subminarr(arr, n, y)
{
    var j = 0;
 
    var stk = [];
 
    // ith index of minarr array
    // will be the index upto which
    // Arr[i] is minimum
    var minarr = Array(n);
    stk.push(0);
 
    for (var i = 1; i < n; i++) {
 
        // Stack is used to find the
        // next smaller element and
        // keeping track of index of
        // current iteration
        while (stk.length!=0
            && arr[i] < arr[stk[stk.length-1]]) {
 
            minarr[stk[stk.length-1]] = i;
            stk.pop();
        }
        stk.push(i);
    }
 
    // Loop for remaining indexes
    while (stk.length!=0) {
 
        minarr[stk[stk.length-1]] = n;
        stk.pop();
    }
 
    var submin = [];
 
    for (var i = 0; i <= n - y; i++) {
 
        // j < i used to keep track
        // whether jth element is inside
        // or outside the window
 
        while (minarr[j] <= i + y - 1
            || j < i) {
            j++;
        }
 
        submin.push(arr[j]);
    }
 
    // Return submin
    return submin;
}
 
// Function to get minimum difference
function getMinDifference(Arr, N, Y)
{
    // Create submin and submax arrays
    var submin
        = get_subminarr(Arr, N, Y);
 
    var submax
        = get_submaxarr(Arr, N, Y);
 
    // Store initial difference
    var minn = submax[0] - submin[0];
    var b = submax.length;
 
    for (var i = 1; i < b; i++) {
 
        // Calculate temporary difference
        var dif = submax[i] - submin[i];
        minn = Math.min(minn, dif);
    }
 
    // Final minimum difference
    document.write( minn + "<br>");
}
 
// Driver Code
// Given array arr[]
var arr = [1, 2, 3, 3, 2, 2];
var N = arr.length
// Given subarray size
var Y = 4;
// Function Call
getMinDifference(arr, N, Y);
 
 
</script>

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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