Given an array arr[], the task is to count the number of times the minimum and the maximum value is updated during the traversal of the array.
Examples:
Input: arr[] = {10, 5, 20, 22}
Output:
Number of times minimum value updated = 2
Number of times maximum value updated = 3
Explanation:
Step 1: Minimum = 10, Maximum = 10
Step 2: Minimum = 5, Maximum = 10
Step 3: Minimum = 5, Maximum = 20
Step 3: Minimum = 5, Maximum = 22
Input: arr[] = {1, 2, 3, 4, 5}
Output:
Number of times minimum value updated = 1
Number of times maximum value updated = 5
Explanation:
Step 1: Minimum = 1, Maximum = 1
Step 2: Minimum = 1, Maximum = 2
Step 3: Minimum = 1, Maximum = 3
Step 4: Minimum = 1, Maximum = 4
Step 5: Minimum = 1, Maximum = 5
Approach: The idea is to keep track of the minimum value and maximum value. Initially initialize these values as the first element of the array. Finally, Iterate over the array, and whenever the maximum or minimum value is to be changed increment the count accordingly.
if (cur_min > arr[i])
cur_min = arr[i]
count_min++;
if (cur_max < arr[i])
cur_max = arr[i]
count_max++;
Below is the implementation of the above approach:
C++
// C++ implementation to find the // number of times minimum and // maximum value updated during the// traversal of the array#include<bits/stdc++.h>using namespace std;// Function to find the number of// times minimum and maximum value// updated during the traversal// of the given arrayvoid maxUpdated(vector<int> arr){ int h_score = arr[0]; int l_score = arr[0]; int i = 1, j = 1; // Increment i if new // highest value occurs // Increment j if new // lowest value occurs for (auto n : arr) { if (h_score < n) { h_score = n; i++; } if (l_score > n) { l_score = n; j++; } } cout << "Number of times maximum value "; cout << "updated = " << i << endl; cout << "Number of times minimum value "; cout << "updated = " << j << endl;}// Driver Codeint main(){ vector<int> arr({10, 5, 20, 22}); maxUpdated(arr);}// This code is contributed by bgangwar59 |
Java
// Java implementation to find the // number of times minimum and // maximum value updated during the// traversal of the arraypublic class GFG { // Function to find the number of // times minimum and maximum value // updated during the traversal // of the given array static void maxUpdated(int[] arr) { int h_score = arr[0]; int l_score = arr[0]; int i = 1, j = 1; // Increment i if new // highest value occurs // Increment j if new // lowest value occurs for (Integer n : arr) { if (h_score < n) { h_score = n; i++; } if (l_score > n) { l_score = n; j++; } } System.out.print( "Number of times maximum value "); System.out.print( "updated = " + i + "\n"); System.out.print( "Number of times minimum value "); System.out.print( "updated = " + j); } // Driver Code public static void main(String[] args) { int[] arr = { 10, 5, 20, 22 }; maxUpdated(arr); }} |
Python
# Python implementation to count # the number of times maximum # and minimum value updated# Function to find the count # the number of times maximum# and minimum value updateddef maximumUpdates(arr): min = arr[0] max = arr[0] minCount, maxCount = 1, 1 # Update the maximum and # minimum values during traversal for arr in arr : if arr>max: maxCount+= 1 max = arr if arr<min: minCount+= 1; min = arr print("Number of times maximum ", end = "") print("value updated = ", maxCount) print("Number of times minimum ", end = "") print("value updated = ", minCount) # Driver code if __name__ == "__main__" : arr = [ 10, 5, 20, 22 ] maximumUpdates(arr) |
C#
// C# implementation to find the // number of times minimum and // maximum value updated during the// traversal of the arrayusing System;class GFG { // Function to find the number of// times minimum and maximum value// updated during the traversal// of the given arraystatic void maxUpdated(int[] arr){ int h_score = arr[0]; int l_score = arr[0]; int i = 1, j = 1; // Increment i if new highest // value occurs Increment j // if new lowest value occurs foreach(int n in arr) { if (h_score < n) { h_score = n; i++; } if (l_score > n) { l_score = n; j++; } } Console.Write("Number of times maximum value "); Console.Write("updated = " + i + "\n"); Console.Write("Number of times minimum value "); Console.Write("updated = " + j);}// Driver Codepublic static void Main(String[] args){ int[] arr = { 10, 5, 20, 22 }; maxUpdated(arr);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript implementation to find the// number of times minimum and// maximum value updated during the// traversal of the array // Function to find the number of // times minimum and maximum value // updated during the traversal // of the given array function maxUpdated(arr) { let h_score = arr[0]; let l_score = arr[0]; let i = 1, j = 1; // Increment i if new // highest value occurs // Increment j if new // lowest value occurs for (let n in arr) { if (h_score < arr[n]) { h_score = n; i++; } if (l_score > n) { l_score = n; j++; } } document.write( "Number of times maximum value "); document.write( "updated = " + i + "<br/>"); document.write( "Number of times minimum value "); document.write( "updated = " + j); }// Driver Code let arr = [ 10, 5, 20, 22 ]; maxUpdated(arr); </script> |
Output:
Number of times maximum value updated = 3 Number of times minimum value updated = 2
Time Complexity: O(n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
