Smallest index in the given array that satisfies the given condition
Given an array arr[] of size N and an integer K, the task is to find the smallest index in the array such that:
floor(arr[i] / 1) + floor(arr[i + 1] / 2) + floor(arr[i + 2] / 3) + ….. + floor(arr[n – 1] / n – i ) ? K
If no such index is found then print -1.
Examples:
Input: arr[] = {6, 5, 4, 2}, K = 8
Output: 1
(6 / 1) + (5 / 2) + (4 / 3) + (2 / 4) = 6 + 2 + 1 + 0 = 9 which is > K
(5 / 1) + (4 / 2) + (2 / 3) = 5 + 2 + 0 = 7 < K
Hence i = 1 is the required index.
Input: arr[] = {5, 4, 2, 3, 9, 1, 8, 7}, K = 7
Output: 5
Approach: For every index i starting from 0, find the sum of the given series, and the first index which given a sum greater than or equal to K is our required index. If there is no such index then print -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the required index int getIndex(int arr[], int n, int K) { // Check all the indices, the first index // satisfying the condition is // the required index for (int i = 0; i < n; i++) { // To store the sum of the series int sum = 0; // Denominator for the sum series int den = 1; // Find the sum of the series for (int j = i; j < n; j++) { sum += (arr[j] / den); // Increment the denominator den++; // If current sum is greater than K // no need to execute loop further if (sum > K) break; } // Found the first valid index if (sum <= K) return i; } return -1; } // Driver code int main() { int arr[] = { 6, 5, 4, 2 }; int n = sizeof(arr) / sizeof(arr[0]); int K = 8; cout << getIndex(arr, n, K); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the required index static int getIndex(int arr[], int n, int K) { // Check all the indices, the first index // satisfying the condition is // the required index for (int i = 0; i < n; i++) { // To store the sum of the series int sum = 0; // Denominator for the sum series int den = 1; // Find the sum of the series for (int j = i; j < n; j++) { sum += (arr[j] / den); // Increment the denominator den++; // If current sum is greater than K // no need to execute loop further if (sum > K) break; } // Found the first valid index if (sum <= K) return i; } return -1; } // Driver code public static void main(String[] args) { int arr[] = { 6, 5, 4, 2 }; int n = arr.length; int K = 8; System.out.print(getIndex(arr, n, K)); } } |
Python
# Python3 implementation of the approach def getIndex(array, k): n = len(array) for ind in range(n): sum = 0 div = 1 for item in array: sum += item//div div += 1 if sum > k: break if sum <= k: return ind return -1 # Driver code arr = [6, 5, 4, 2] K = 8print(getIndex(arr, K)) arr = [5, 4, 2, 3, 9, 1, 8, 7] K = 7print(getIndex(arr, K)) |
C#
// C# implementation of the approach using System; class GFG { // Function to return the required index static int getIndex(int []arr, int n, int K) { // Check all the indices, the first index // satisfying the condition is // the required index for (int i = 0; i < n; i++) { // To store the sum of the series int sum = 0; // Denominator for the sum series int den = 1; // Find the sum of the series for (int j = i; j < n; j++) { sum += (arr[j] / den); // Increment the denominator den++; // If current sum is greater than K // no need to execute loop further if (sum > K) break; } // Found the first valid index if (sum <= K) return i; } return -1; } // Driver code static public void Main () { int []arr = { 6, 5, 4, 2 }; int n = arr.Length; int K = 8; Console.WriteLine(getIndex(arr, n, K)); } } // This Code is contributed by ajit. |
PHP
<?php // PHP implementation of the approach // Function to return the required index function getIndex($arr, $n, $K) { // Check all the indices, the first // index satisfying the condition is // the required index for ($i = 0; $i < $n; $i++) { // To store the sum of the series $sum = 0; // Denominator for the sum series $den = 1; // Find the sum of the series for ($j = $i; $j < $n; $j++) { $sum += floor($arr[$j] / $den); // Increment the denominator $den++; // If current sum is greater than K // no need to execute loop further if ($sum > $K) break; } // Found the first valid index if ($sum <= $K) return $i; } return -1; } // Driver code $arr = array( 6, 5, 4, 2 ); $n = sizeof($arr); $K = 8; echo getIndex($arr, $n, $K); // This code is contributed by Ryuga ?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the required index function getIndex(arr, n, K) { // Check all the indices, the first index // satisfying the condition is // the required index for (let i = 0; i < n; i++) { // To store the sum of the series let sum = 0; // Denominator for the sum series let den = 1; // Find the sum of the series for (let j = i; j < n; j++) { sum += parseInt((arr[j] / den), 10); // Increment the denominator den++; // If current sum is greater than K // no need to execute loop further if (sum > K) break; } // Found the first valid index if (sum <= K) return i; } return -1; } let arr = [ 6, 5, 4, 2 ]; let n = arr.length; let K = 8; document.write(getIndex(arr, n, K)); </script> |
Output:
1
Time complexity: O(N2)
Auxiliary space: O(1)
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