Skip to content
Related Articles

Related Articles

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

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 11 Jul, 2022

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 ith index to (i + Y – 1)th index 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 ith 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:

  1. Build two arrays maxarr[] and minarr[], where maxarr[] will store the index of the element which is next greater to the element at ith index and minarr[] will store the index of the next element which is less than the element at the ith index.
  2. Initialize a stack with 0 to store the indices in both the above cases.
  3. 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 ith iteration.
  4. 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; 
}

Java




// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to get the maximum of all
// the subarrays of size Y
static Vector<Integer> get_submaxarr(int[] arr, int n, 
                                     int y)
{
    int j = 0;
    Stack<Integer> stk = new Stack<Integer>();
    
    // ith index of maxarr array
    // will be the index upto which
    // Arr[i] is maximum
    int[] maxarr = new int[n];
    Arrays.fill(maxarr,0);
    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.size() != 0 && 
               arr[i] > arr[stk.peek()]) 
        {
            maxarr[stk.peek()] = i - 1;
            stk.pop();
        }
        stk.push(i);
    }
    
    // Loop for remaining indexes
    while (stk.size() != 0)
    {
        maxarr[stk.size()] = n - 1;
        stk.pop();
    }
    Vector<Integer> submax = new Vector<Integer>();
    
    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 || j < i) 
        {
            j++;
        }
        submax.add(arr[j]);
    }
    
    // Return submax
    return submax;
}
    
// Function to get the minimum for
// all subarrays of size Y
static Vector<Integer> get_subminarr(int[] arr, int n,
                                     int y)
{
    int j = 0;
    
    Stack<Integer> stk = new Stack<Integer>();
    
    // ith index of minarr array
    // will be the index upto which
    // Arr[i] is minimum
    int[] minarr = new int[n];
    Arrays.fill(minarr,0);
    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.size() != 0 && 
               arr[i] < arr[stk.peek()])
        {
            minarr[stk.peek()] = i;
            stk.pop();
        }
        stk.push(i);
    }
    
    // Loop for remaining indexes
    while (stk.size() != 0)
    {
        minarr[stk.peek()] = n;
        stk.pop();
    }
    
    Vector<Integer> submin = new Vector<Integer>();
    
    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 || j < i) 
        {
            j++;
        }
        submin.add(arr[j]);
    }
    
    // Return submin
    return submin;
}
    
// Function to get minimum difference
static void getMinDifference(int[] Arr, int N, int Y)
{
      
    // Create submin and submax arrays
    Vector<Integer> submin = get_subminarr(Arr, N, Y);
    
    Vector<Integer> submax = get_submaxarr(Arr, N, Y);
    
    // Store initial difference
    int minn = submax.get(0) - submin.get(0);
    int b = submax.size();
    
    for(int i = 1; i < b; i++) 
    {
          
        // Calculate temporary difference
        int dif = submax.get(i) - submin.get(i) + 1;
        minn = Math.min(minn, dif);
    }
    
    // Final minimum difference
    System.out.print(minn);
}
  
// Driver code
public static void main(String[] args)
{
      
    // Given array arr[]
    int[] arr = { 1, 2, 3, 3, 2, 2 };
    int N = arr.length;
    
    // Given subarray size
    int Y = 4;
    
    // Function Call
    getMinDifference(arr, N, Y);
}
}
  
// This code is contributed by decode2207

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] *
    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

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
      
    // Function to get the maximum of all
    // the subarrays of size Y
    static List<int> get_submaxarr(int[] arr, int n, int y)
    {
        int j = 0;
        Stack<int> stk = new Stack<int>();
       
        // ith index of maxarr array
        // will be the index upto which
        // Arr[i] is maximum
        int[] maxarr = new int[n];
        Array.Fill(maxarr,0);
        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.Count!=0 && arr[i] > arr[stk.Peek()]) {
       
                maxarr[stk.Peek()] = i - 1;
                stk.Pop();
            }
            stk.Push(i);
        }
       
        // Loop for remaining indexes
        while (stk.Count!=0) {
       
            maxarr[stk.Count] = n - 1;
            stk.Pop();
        }
        List<int> submax = new List<int>();
       
        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
                || j < i) {
                j++;
            }
       
            submax.Add(arr[j]);
        }
       
        // Return submax
        return submax;
    }
       
    // Function to get the minimum for
    // all subarrays of size Y
    static List<int> get_subminarr(int[] arr, int n, int y)
    {
        int j = 0;
       
        Stack<int> stk = new Stack<int>();
       
        // ith index of minarr array
        // will be the index upto which
        // Arr[i] is minimum
        int[] minarr = new int[n];
        Array.Fill(minarr,0);
        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.Count!=0
                && arr[i] < arr[stk.Peek()]) {
       
                minarr[stk.Peek()] = i;
                stk.Pop();
            }
            stk.Push(i);
        }
       
        // Loop for remaining indexes
        while (stk.Count!=0) {
       
            minarr[stk.Peek()] = n;
            stk.Pop();
        }
       
        List<int> submin = new List<int>();
       
        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
                || j < i) {
                j++;
            }
       
            submin.Add(arr[j]);
        }
       
        // Return submin
        return submin;
    }
       
    // Function to get minimum difference
    static void getMinDifference(int[] Arr, int N, int Y)
    {
        // Create submin and submax arrays
        List<int> submin
            = get_subminarr(Arr, N, Y);
       
        List<int> submax
            = get_submaxarr(Arr, N, Y);
       
        // Store initial difference
        int minn = submax[0] - submin[0];
        int b = submax.Count;
       
        for (int i = 1; i < b; i++) {
       
            // Calculate temporary difference
            int dif = submax[i] - submin[i] + 1;
            minn = Math.Min(minn, dif);
        }
       
        // Final minimum difference
        Console.WriteLine(minn);
    }
  
  static void Main() {
    // Given array arr[]
    int[] arr = { 1, 2, 3, 3, 2, 2 };
    int N = arr.Length;
   
    // Given subarray size
    int Y = 4;
   
    // Function Call
    getMinDifference(arr, N, Y);
  }
}
  
// This code is contributed by rameshtravel07.

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: 

1

 

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

Related Topic: Subarrays, Subsequences, and Subsets in Array


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!