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 answer
question 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 ans
C++
// 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 k 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 (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 program int 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 element public 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 k def calculateMaxSumLength(arr, n, k): ans = 0 # final sum of lengths for i in range (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 Program arr = [ 4 , 5 , 7 , 1 , 2 , 9 , 8 , 4 , 3 , 1 ] size = len (arr) k = 4 ans = calculateMaxSumLength(arr, size, k) print "Max Length ::" ,ans # Contributed by Harshit Agrawal |
C#
// A C# program to calculate max // sum lengths of non overlapping // contiguous subarrays with k as // max element using 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 k function 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 ?> |
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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