Given two integers X, Y and a range [L, R], the task is to count the number of integers from the range X and Y (inclusive) that can be visited in any number of steps, starting from X. In each step, it is possible to increase by L to R.
Examples:
Input: X = 1, Y = 10, L = 4, R = 6
Output: 6
Explanation: A total of six points can be visited between [1, 10]:
- 1: Starting point
- 5: 1 -> 5
- 6: 1 -> 6
- 7: 1 -> 7
- 9: 1 -> 5 -> 9
- 10: 1 -> 5 -> 10
Input: X = 3, Y = 12, L = 2, R = 3
Output: 9
Approach: From index i, one can reach anywhere between [i+L, i+R] also similarly for each point j in [i+L, i+R] one can move to [j+L, j+R]. For each index i these reachable ranges can be marked using the difference array. Follow the steps below for the approach.
- Construct an array say diff_arr[] of large size, to mark the ranges.
- Initialize a variable say count = 0, to count the reachable points.
- Initially, make diff_arr[x] = 1 and diff_arr[x+1] = -1 to mark starting point X as visited.
- Iterate from X to Y and for each i update diff_arr[i] += diff_arr[i-1] and if diff_arr[i] >= 1 then update diff_arr[i+L] = 1 and diff_arr[i + R + 1] = -1 and count = count + 1.
- Finally, return the count.
Below is the implementation of the above approach:
C++
// C++ code for above approach#include <bits/stdc++.h>using namespace std;// Function to count points from the range [X, Y]// that can be reached by moving by L or R stepsint countReachablePoints(int X, int Y, int L, int R){ // Initialize difference array int diff_arr[100000] = { 0 }; // Initialize Count int count = 0; // Marking starting point diff_arr[X] = 1; diff_arr[X + 1] = -1; // Iterating from X to Y for (int i = X; i <= Y; i++) { // Accumulate difference array diff_arr[i] += diff_arr[i - 1]; // If diff_arr[i] is greater // than 1 if (diff_arr[i] >= 1) { // Updating difference array diff_arr[i + L] += 1; diff_arr[i + R + 1] -= 1; // Visited point found count++; } } return count;}// Driver Codeint main(){ // Given Input int X = 3, Y = 12, L = 2, R = 3; // Function Call cout << countReachablePoints(X, Y, L, R); return 0;} |
Java
// Java code for above approachimport java.util.*;class GFG{// Function to count points from the range [X, Y]// that can be reached by moving by L or R stepsstatic int countReachablePoints(int X, int Y, int L, int R){ // Initialize difference array int diff_arr[] = new int[100000]; // Initialize Count int count = 0; // Marking starting point diff_arr[X] = 1; diff_arr[X + 1] = -1; // Iterating from X to Y for (int i = X; i <= Y; i++) { // Accumulate difference array diff_arr[i] += diff_arr[i - 1]; // If diff_arr[i] is greater // than 1 if (diff_arr[i] >= 1) { // Updating difference array diff_arr[i + L] += 1; diff_arr[i + R + 1] -= 1; // Visited point found count++; } } return count;}// Driver Codepublic static void main(String[] args){ // Given Input int X = 3, Y = 12, L = 2, R = 3; // Function Call System.out.print(countReachablePoints(X, Y, L, R));}}// This code contributed by shikhasingrajput |
Python3
# Python3 code for above approach# Function to count points from the range [X, Y]# that can be reached by moving by L or R stepsdef countReachablePoints(X, Y, L, R): # Initialize difference array diff_arr = [0 for i in range(100000)] # Initialize Count count = 0 # Marking starting point diff_arr[X] = 1 diff_arr[X + 1] = -1 # Iterating from X to Y for i in range(X, Y + 1, 1): # Accumulate difference array diff_arr[i] += diff_arr[i - 1] # If diff_arr[i] is greater # than 1 if (diff_arr[i] >= 1): # Updating difference array diff_arr[i + L] += 1 diff_arr[i + R + 1] -= 1 # Visited point found count += 1 return count# Driver Codeif __name__ == '__main__': # Given Input X = 3 Y = 12 L = 2 R = 3 # Function Call print(countReachablePoints(X, Y, L, R)) # This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;class GFG{// Function to count points from the range [X, Y]// that can be reached by moving by L or R stepsstatic int countReachablePoints(int X, int Y, int L, int R){ // Initialize difference array int[] diff_arr= new int[100000]; // Initialize Count int count = 0; // Marking starting point diff_arr[X] = 1; diff_arr[X + 1] = -1; // Iterating from X to Y for (int i = X; i <= Y; i++) { // Accumulate difference array diff_arr[i] += diff_arr[i - 1]; // If diff_arr[i] is greater // than 1 if (diff_arr[i] >= 1) { // Updating difference array diff_arr[i + L] += 1; diff_arr[i + R + 1] -= 1; // Visited point found count++; } } return count;}// Driver Codepublic static void Main(){ // Given Input int X = 3, Y = 12, L = 2, R = 3; // Function Call Console.Write(countReachablePoints(X, Y, L, R));}}// This code is contributed by splevel62. |
Javascript
<script>// JavaScript code for above approach// Function to count points from the range [X, Y]// that can be reached by moving by L or R stepsfunction countReachablePoints(X, Y, L, R) { // Initialize difference array let diff_arr = new Array(100000).fill(0); // Initialize Count let count = 0; // Marking starting point diff_arr[X] = 1; diff_arr[X + 1] = -1; // Iterating from X to Y for (let i = X; i <= Y; i++) { // Accumulate difference array diff_arr[i] += diff_arr[i - 1]; // If diff_arr[i] is greater // than 1 if (diff_arr[i] >= 1) { // Updating difference array diff_arr[i + L] += 1; diff_arr[i + R + 1] -= 1; // Visited point found count++; } } return count;}// Driver Code// Given Inputlet X = 3, Y = 12, L = 2, R = 3;// Function Calldocument.write(countReachablePoints(X, Y, L, R));</script> |
Output:
9
Time Complexity: O(Y – X)
Auxiliary Space: O(Y)
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