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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 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;
}

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: 
1

 

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

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