Given a range from L to R and an integer K, the task is to count the number of integers in the given range such that their last K digits are equal.
Example:
Input: L = 49, R = 101, K=2
Output: 6
Explanation: There are 6 possible integers t.e., 55, 66, 77, 88, 99 and 100 such that their last K(i.e., 2) digits are equal.Input: L = 10, R = 20, K=2
Output: 1
Efficient Approach: It can be observed that the count of integers i in the range 1 to X having the last K digits equal to an integer z (i.e., i % 10K = z) are (X – z)/10K + 1. Using this observation the above problem can be solved using the below steps:
- Suppose intCount(X, K) represents the count of integers from 1 to X having the last K digits as equal.
- To calculate intCount(X, K), iterate over all possibilities of z having K digits (i.e., {00…0, 11…1, 22…2, 33…3, 44…4, …., 99…9 }) in the formula (X – z)/10K +1 and maintain their sum which is the required value.
- Therefore, the count of integers in range L to R can be obtained as intCount(R, K) – intCount(L-1, K).
Below is the implementation of the above approach:
C++
// C++ Program of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the count of// integers from 1 to X having the// last K digits as equalint intCount(int X, int K){ // Stores the total count of integers int ans = 0; // Loop to iterate over all // possible values of z for (int z = 0; z < pow(10, K); z += (pow(10, K) - 1) / 9) { // Terminate the loop when z > X if (z > X) break; // Add count of integers with // last K digits equal to z ans += ((X - z) / pow(10, K) + 1); } // Return count return ans;}// Function to return the count of// integers from L to R having the// last K digits as equalint intCountInRange(int L, int R, int K){ return (intCount(R, K) - intCount(L - 1, K));}// Driver Codeint main(){ int L = 49; int R = 101; int K = 2; // Function Call cout << intCountInRange(L, R, K); return 0;} |
Java
// Java Program of the above approachimport java.util.*;class GFG{// Function to return the count of// integers from 1 to X having the// last K digits as equalstatic int intCount(int X, int K){ // Stores the total count of integers int ans = 0; // Loop to iterate over all // possible values of z for (int z = 0; z < Math.pow(10, K); z += (Math.pow(10, K) - 1) / 9) { // Terminate the loop when z > X if (z > X) break; // Add count of integers with // last K digits equal to z ans += ((X - z) / Math.pow(10, K) + 1); } // Return count return ans;}// Function to return the count of// integers from L to R having the// last K digits as equalstatic int intCountInRange(int L, int R, int K){ return (intCount(R, K) - intCount(L - 1, K));}// Driver Codepublic static void main(String[] args){ int L = 49; int R = 101; int K = 2; // Function Call System.out.print(intCountInRange(L, R, K));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to return the count of# integers from 1 to X having the# last K digits as equaldef intCount(X, K): # Stores the total count of integers ans = 0 # Loop to iterate over all # possible values of z for z in range(0, int(pow(10, K)), int((pow(10, K) - 1) / 9)): # Terminate the loop when z > X if (z > X): break # Add count of integers with # last K digits equal to z ans += int((X - z) / int(pow(10, K)) + 1) # Return count return ans# Function to return the count of# integers from L to R having the# last K digits as equaldef intCountInRange(L, R, K): return(intCount(R, K) - intCount(L - 1, K))# Driver CodeL = 49R = 101K = 2# Function Callprint(intCountInRange(L, R, K))# This code is contributed by sanjoy_62 |
C#
// C# Program of the above approachusing System;class GFG{// Function to return the count of// integers from 1 to X having the// last K digits as equalstatic int intCount(int X, int K){ // Stores the total count of integers int ans = 0; // Loop to iterate over all // possible values of z for (int z = 0; z < Math.Pow(10, K); z += ((int)Math.Pow(10, K) - 1) / 9) { // Terminate the loop when z > X if (z > X) break; // Add count of integers with // last K digits equal to z ans += ((X - z) / (int)Math.Pow(10, K) + 1); } // Return count return ans;}// Function to return the count of// integers from L to R having the// last K digits as equalstatic int intCountInRange(int L, int R, int K){ return (intCount(R, K) - intCount(L - 1, K));}// Driver Codepublic static void Main(String[] args){ int L = 49; int R = 101; int K = 2; // Function Call Console.Write(intCountInRange(L, R, K));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// JavaScript program for the above approach// Function to return the count of// integers from 1 to X having the// last K digits as equalfunction intCount(X, K){ // Stores the total count of integers let ans = 0; // Loop to iterate over all // possible values of z for(let z = 0; z < Math.pow(10, K); z += Math.floor((Math.pow(10, K) - 1) / 9)) { // Terminate the loop when z > X if (z > X) break; // Add count of integers with // last K digits equal to z ans += Math.floor(((X - z) / Math.pow(10, K) + 1)); } // Return count return ans;}// Function to return the count of// integers from L to R having the// last K digits as equalfunction intCountInRange(L, R, K){ return(intCount(R, K) - intCount(L - 1, K));}// Driver Codelet L = 49;let R = 101;let K = 2;// Function Calldocument.write(intCountInRange(L, R, K));// This code is contributed by Potta Lokesh</script> |
Output
6
Time Complexity: O(log K)
Space Complexity: O(1)
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