Maximum sum of lengths of non-overlapping subarrays with k as the max element.
Find the maximum sum of lengths of non-overlapping subarrays (contiguous elements) with k as the maximum element.
Examples:
Input : arr[] = {2, 1, 4, 9, 2, 3, 8, 3, 4}
k = 4
Output : 5
{2, 1, 4} => Length = 3
{3, 4} => Length = 2
So, 3 + 2 = 5 is the answer
Input : arr[] = {1, 2, 3, 2, 3, 4, 1}
k = 4
Output : 7
{1, 2, 3, 2, 3, 4, 1} => Length = 7
Input : arr = {4, 5, 7, 1, 2, 9, 8, 4, 3, 1}
k = 4
Ans = 4
{4} => Length = 1
{4, 3, 1} => Length = 3
So, 1 + 3 = 4 is the answerquestion source : https://www.geeksforgeeks.org/amazon-interview-experience-set-376-campus-internship/
Algorithm :
Traverse the array starting from first element
Take a loop and keep on incrementing count
If element is less than equal to k
if array element is equal to k, then mark
a flag
If flag is marked, add this count to answer
Take another loop and traverse the array
till element is greater than k
return ansC++
// CPP program to calculate max sum lengths of// non overlapping contiguous subarrays with k as// max element#include <bits/stdc++.h>using namespace std; // Returns max sum of lengths with maximum element// as kint calculateMaxSumLength(int arr[], int n, int k){ int ans = 0; // final sum of lengths // number of elements in current subarray int count = 0; // variable for checking if k appeared in subarray int flag = 0; for (int i = 0; i < n;) { count = 0; flag = 0; // count the number of elements which are // less than equal to k while (arr[i] <= k && i < n) { count++; if (arr[i] == k) flag = 1; i++; } // if current element appeared in current // subarray add count to sumLength if (flag == 1) ans += count; // skip the array elements which are // greater than k while (arr[i] > k && i < n) i++; } return ans;} // driver programint main(){ int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 }; int size = sizeof(arr) / sizeof(arr[0]); int k = 4; int ans = calculateMaxSumLength(arr, size, k); cout << "Max Length :: " << ans << endl; return 0;} |
Java
// A Java program to calculate max sum lengths of// non overlapping contiguous subarrays with k as// max elementpublic class GFG{ // Returns max sum of lengths with maximum element // as k static int calculateMaxSumLength(int arr[], int n, int k) { int ans = 0; // final sum of lengths // number of elements in current subarray int count = 0; // variable for checking if k appeared in subarray int flag = 0; for (int i = 0; i < n;) { count = 0; flag = 0; // count the number of elements which are // less than equal to k while (i < n && arr[i] <= k) { count++; if (arr[i] == k) flag = 1; i++; } // if current element appeared in current // subarray add count to sumLength if (flag == 1) ans += count; // skip the array elements which are // greater than k while (i < n && arr[i] > k) i++; } return ans; } // driver program to test above method public static void main(String[] args) { int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 }; int size = arr.length; int k = 4; int ans = calculateMaxSumLength(arr, size, k); System.out.println("Max Length :: " + ans); }}// This code is contributed by Sumit Ghosh |
Python
# Python program to calculate max sum lengths of non# overlapping contiguous subarrays with k as max element# Returns max sum of lengths with max elements as kdef calculateMaxSumLength(arr, n, k): ans = 0 # final sum of lengths i=0 while i < n : # number of elements in current sub array count = 0 # Variable for checking if k appeared in the sub array flag = 0 # Count the number of elements which are # less than or equal to k while i < n and arr[i] <= k : count = count + 1 if arr[i] == k: flag = 1 i = i + 1 # if current element appeared in current # subarray and count to sumLength if flag == 1: ans = ans + count # skip the array elements which are greater than k while i < n and arr[i] > k : i = i + 1 return ans # Driver Programarr = [4, 5, 7, 1, 2, 9, 8, 4, 3, 1]size = len(arr)k = 4ans = calculateMaxSumLength(arr, size, k)print "Max Length ::",ans# Contributed by Rohit |
C#
// A C# program to calculate max// sum lengths of non overlapping// contiguous subarrays with k as// max elementusing System;class GFG { // Returns max sum of lengths // with maximum element as k static int calculateMaxSumLength(int []arr, int n, int k) { // final sum of lengths int ans = 0; // number of elements in // current subarray int count = 0; // variable for checking if // k appeared in subarray int flag = 0; for(int i = 0; i < n;) { count = 0; flag = 0; // count the number of // elements which are // less than equal to k while (i < n && arr[i] <= k) { count++; if (arr[i] == k) flag = 1; i++; } // if current element // appeared in current // subarray add count // to sumLength if (flag == 1) ans += count; // skip the array // elements which are // greater than k while (i < n && arr[i] > k) i++; } return ans; } // Driver Code public static void Main() { int []arr = {4, 5, 7, 1, 2, 9, 8, 4, 3, 1}; int size = arr.Length; int k = 4; int ans = calculateMaxSumLength(arr, size, k); Console.WriteLine("Max Length :: " + ans); }}// This code is contributed by anuj_67. |
PHP
<?php// PHP program to calculate max sum lengths// of non overlapping contiguous subarrays// with k as max element// Returns max sum of lengths with maximum// element as kfunction calculateMaxSumLength(&$arr, $n, $k){ $ans = 0; // final sum of lengths // number of elements in current subarray $count = 0; // variable for checking if k // appeared in subarray $flag = 0; for ($i = 0; $i < $n😉 { $count = 0; $flag = 0; // count the number of elements which // are less than equal to k while ($arr[$i] <= $k && $i < $n) { $count++; if ($arr[$i] == $k) $flag = 1; $i++; } // if current element appeared in current // subarray add count to sumLength if ($flag == 1) $ans += $count; // skip the array elements which are // greater than k while ($arr[$i] > $k && $i < $n) $i++; } return $ans;}// Driver Code$arr = array( 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 );$size = sizeof($arr);$k = 4;$ans = calculateMaxSumLength($arr, $size, $k);echo "Max Length :: " . $ans . "\n";// This code is contributed by ita_c?> |
Javascript
<script>// A Javascript program to calculate max sum lengths of// non overlapping contiguous subarrays with k as// max element // Returns max sum of lengths with maximum element // as k function calculateMaxSumLength(arr,n,k) { let ans = 0; // final sum of lengths // number of elements in current subarray let count = 0; // variable for checking if k appeared in subarray let flag = 0; for (let i = 0; i < n;) { count = 0; flag = 0; // count the number of elements which are // less than equal to k while (i < n && arr[i] <= k) { count++; if (arr[i] == k) flag = 1; i++; } // if current element appeared in current // subarray add count to sumLength if (flag == 1) ans += count; // skip the array elements which are // greater than k while (i < n && arr[i] > k) i++; } return ans; } // driver program to test above method let arr=[4, 5, 7, 1, 2, 9, 8, 4, 3, 1]; let size = arr.length; let k = 4; let ans = calculateMaxSumLength(arr, size, k); document.write("Max Length :: " + ans); //This code is contributed by avanitrachhadiya2155 </script> |
Output:
Max Length :: 4
Time Complexity : O(n)
It may look like O(n2), but if you take a closer look, array is traversed only once
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