Given a very large number, print all the 3 digit repeating numbers with their frequency. If a 3 digit number appears more than once, print the number and its frequency.
Example:
Input: 123412345123456
Output: 123 - 3 times
234 - 3 times
345 - 2 times
Input: 43243243
Output: 432 - 2 times
324 - 2 times
243 - 2 times
Approach: Since the number is very large, it is stored in a string. Initially, the first three-digit number will be the first three characters from the left. Iterate in the string from the 3rd index from the left in the string and do %100 to remove the first character and append the ith index number at the end to get the new number. Increase the frequency of the number in the hash map. In the end, when all the 3-digit numbers are generated, print all the numbers which have a frequency of more than 1.
Below is the implementation of the above idea:
C++
// CPP program to print 3 digit repeating numbers#include <bits/stdc++.h>using namespace std;// function to print 3// digit repeating numbersvoid printNum(string s){ int i = 0, j = 0, val = 0; // Hashmap to store the // frequency of a 3 digit number map <int, int> mp; // first three digit number val = (s[0] - '0') * 100 + (s[1] - '0') * 10 + (s[2] - '0'); mp[val] = 1; for (i = 3; i < s.length(); i++) { val = (val % 100) * 10 + s[i] - '0'; // if key already exists // increase value by 1 if (mp.find(val) != mp.end()) { mp[val] = mp[val] + 1; } else { mp[val] = 1; } } // Output the three digit numbers with frequency>1 for (auto m : mp) { int key = m.first; int value = m.second; if (value > 1) cout << key << " - " << value << " times" << endl; }}// Driver Codeint main() { // Input string string input = "123412345123456"; // Calling Function printNum(input);}// This code is contributed by Nishant Tanwar |
Java
// Java program to print 3 digit repeating numbersimport java.util.*;import java.lang.*;public class GFG { // function to print 3 // digit repeating numbers static void printNum(String s) { int i = 0, j = 0, val = 0; // Hashmap to store the // frequency of a 3 digit number LinkedHashMap<Integer, Integer> hm = new LinkedHashMap<>(); // first three digit number val = (s.charAt(0) - '0') * 100 + (s.charAt(1) - '0') * 10 + (s.charAt(2) - '0'); hm.put(val, 1); for (i = 3; i < s.length(); i++) { val = (val % 100) * 10 + s.charAt(i) - '0'; // if key already exists // increase value by 1 if (hm.containsKey(val)) { hm.put(val, hm.get(val) + 1); } else { hm.put(val, 1); } } // Output the three digit numbers with frequency>1 for (Map.Entry<Integer, Integer> en : hm.entrySet()) { int key = en.getKey(); int value = en.getValue(); if (value > 1) System.out.println(key + " - " + value + " times"); } } // Driver Code public static void main(String args[]) { // Input string String input = "123412345123456"; // Calling Function printNum(input); }} |
Python3
# Python3 program to print # 3 digit repeating numbers# Function to print 3 # digit repeating numbers def printNum(s): i, j, val = 0, 0, 0 # Hashmap to store the # frequency of a 3 digit number mp = {} # first three digit number val = ((ord(s[0]) - ord('0')) * 100 + (ord(s[1]) - ord('0')) * 10 + (ord(s[2]) - ord('0'))) mp[val] = 1 for i in range (3, len(s)): val = (val % 100) * 10 + ord(s[i]) - ord('0') # if key already exists # increase value by 1 if (val in mp): mp[val] = mp[val] + 1 else: mp[val] = 1 # Output the three digit # numbers with frequency>1 for m in mp: key = m value = mp[m] if (value > 1): print (key, " - ", value, " times") # Driver Code if __name__ == "__main__": # Input string input = "123412345123456" # Calling Function printNum(input)# This code is contributed by Chitranayal |
C#
// C# program to print 3 digit repeating numbersusing System;using System.Collections.Generic; class GFG{ // function to print 3 // digit repeating numbers static void printNum(String s) { int i = 0, val = 0; // Hashmap to store the // frequency of a 3 digit number Dictionary<int, int> hm = new Dictionary<int, int>(); // first three digit number val = (s[0] - '0') * 100 + (s[1] - '0') * 10 + (s[2] - '0'); hm.Add(val, 1); for (i = 3; i < s.Length; i++) { val = (val % 100) * 10 + s[i] - '0'; // if key already exists // increase value by 1 if (hm.ContainsKey(val)) { hm[val] = hm[val] + 1; } else { hm.Add(val, 1); } } // Output the three digit numbers with frequency>1 foreach(KeyValuePair<int, int> en in hm) { int key = en.Key; int value = en.Value; if (value > 1) Console.WriteLine(key + " - " + value + " times"); } } // Driver Code public static void Main(String []args) { // Input string String input = "123412345123456"; // Calling Function printNum(input); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to print 3 digit repeating numbers // function to print 3 // digit repeating numbers function printNum(s) { let i = 0, j = 0, val = 0; // Hashmap to store the // frequency of a 3 digit number let hm = new Map(); // first three digit number val = (s[0].charCodeAt(0) - '0'.charCodeAt(0)) * 100 + (s[1].charCodeAt(0) - '0'.charCodeAt(0)) * 10 + (s[2].charCodeAt(0) - '0'.charCodeAt(0)); hm.set(val, 1); for (i = 3; i < s.length; i++) { val = (val % 100) * 10 + s[i].charCodeAt(0) - '0'.charCodeAt(0); // if key already exists // increase value by 1 if (hm.has(val)) { hm.set(val, hm.get(val) + 1); } else { hm.set(val, 1); } } // Output the three digit numbers with frequency>1 for (let [Key, Value] of hm.entries()) { let key = Key; let value = Value; if (value > 1) document.write(key + " - " + value + " times<br>"); } } // Driver Code // Input string let input = "123412345123456"; // Calling Function printNum(input); // This code is contributed by avanitrachhadiya2155</script> |
Output
123 - 3 times 234 - 3 times 345 - 2 times
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