Given an array arr[] of N positive integers and a positive integer K, the task is to find the range [L, R] such that for all the elements in this range, say S each array element arr[i] is floor((i*S)/K).
Examples:
Input: N = 5, K = 10, arr[] = {2, 4, 6, 9, 11}
Output: 23 23
Explanation:
When S = 23, substituting in the equation gives the same array values:
arr[1] = floor((1*23)/10) = 2
arr[2] = floor((2*23/10)) = 4
arr[3] = floor((3*23/10)) = 6
arr[4] = floor((4*23/10)) = 9
arr[5] = floor((5*23/10)) = 11Input: N = 5, K = 100, arr[] = {0, 0, 0, 0, 1}
Output: 20 24
Approach: Follow the below steps to solve the given problem:
- Initialize two variables say L and R as INT_MIN and INT_MAX that stores the value of the left range and the right range respectively.
- Traverse the given array arr[] and perform the following steps:
- Find the possible value of the left range for the ith array element as ceil(1.0*arr[i]*K/(i + 1)).
- Find the possible value of the right range for the ith array element as ceil((1.0 + arr[i])*K/(i + 1)) – 1.
- Update the value of L as max(L, l) and the value of R as min(R, r).
- After completing the above steps, print the values of L and R as the resultant range.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the range of values// for S in a given array that satisfies// the given conditionvoid findRange(int arr[], int N, int K){ // Stores the left range value int L = INT_MIN; // Stores the right range value int R = INT_MAX; for (int i = 0; i < N; i++) { // Find the current left range // value for S int l = (int)ceil(1.0 * arr[i] * K / (i + 1)); // Find the current right range // value for S int r = (int)ceil((1.0 + arr[i]) * K / (i + 1)) - 1; // Updating L value L = max(L, l); // Updating R value R = min(R, r); } cout << L << " " << R;}// Driver Codeint main(){ int arr[] = { 2, 4, 6, 9, 11 }; int K = 10; int N = sizeof(arr) / sizeof(int); findRange(arr, N, K); return 0;}// This code is contributed by Potta Lokesh |
Java
// Java program for the above approachimport java.util.*;import java.lang.*;class Codechef { // Function to find the range of values // for S in a given array that satisfies // the given condition static void findRange(int arr[], int N, int K) { // Stores the left range value int L = Integer.MIN_VALUE; // Stores the right range value int R = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { // Find the current left range // value for S int l = (int)Math.ceil( 1.0 * arr[i] * K / (i + 1)); // Find the current right range // value for S int r = (int)Math.ceil( (1.0 + arr[i]) * K / (i + 1)) - 1; // Updating L value L = Math.max(L, l); // Updating R value R = Math.min(R, r); } System.out.println(L + " " + R); } // Driver Code public static void main(String[] args) { int arr[] = { 2, 4, 6, 9, 11 }; int K = 10; int N = arr.length; findRange(arr, N, K); }} |
Python3
# Python 3 program for the above approachfrom math import ceil,floorimport sys# Function to find the range of values# for S in a given array that satisfies# the given conditiondef findRange(arr, N, K): # Stores the left range value L = -sys.maxsize-1 # Stores the right range value R = sys.maxsize for i in range(N): # Find the current left range # value for S l = ceil(1.0 * arr[i] * K / (i + 1)) # Find the current right range # value for S r = ceil((1.0 + arr[i]) * K / (i + 1) - 1) # Updating L value L = max(L, l) # Updating R value R = min(R, r) print(L,R)# Driver Codeif __name__ == '__main__': arr = [2, 4, 6, 9, 11] K = 10 N = len(arr) findRange(arr, N, K) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;class GFG { // Function to find the range of values // for S in a given array that satisfies // the given condition static void findRange(int[] arr, int N, int K) { // Stores the left range value int L = Int32.MinValue; // Stores the right range value int R = Int32.MaxValue; for (int i = 0; i < N; i++) { // Find the current left range // value for S int l = (int)Math.Ceiling(1.0 * arr[i] * K / (i + 1)); // Find the current right range // value for S int r = (int)Math.Ceiling((1.0 + arr[i]) * K / (i + 1)) - 1; // Updating L value L = Math.Max(L, l); // Updating R value R = Math.Min(R, r); } Console.WriteLine(L + " " + R); } // Driver Code public static void Main() { int[] arr = { 2, 4, 6, 9, 11 }; int K = 10; int N = arr.Length; findRange(arr, N, K); }}// This code is contributed by subham348. |
Javascript
<script>// JavaScript program for the above approach// Function to find the range of values// for S in a given array that satisfies// the given conditionfunction findRange(arr, N, K){ // Stores the left range value let L = Number.MIN_VALUE; // Stores the right range value let R = Number.MAX_VALUE; for (let i = 0; i < N; i++) { // Find the current left range // value for S let l = Math.ceil(1.0 * arr[i] * K / (i + 1)); // Find the current right range // value for S let r = Math.ceil((1.0 + arr[i]) * K / (i + 1)) - 1; // Updating L value L = Math.max(L, l); // Updating R value R = Math.min(R, r); } document.write(L + " " + R);} // Driver code let arr = [ 2, 4, 6, 9, 11 ]; let K = 10; let N = arr.length; findRange(arr, N, K);// This code is contributed by AnkThon</script> |
Output:
23 23
Time Complexity: O(N)
Auxiliary Space: O(1)
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