Find the only element that appears b times
Given an array where every element occurs a times, except one element which occurs b (a>b) times. Find the element that occurs b times.
Examples:
Input : arr[] = [1, 1, 2, 2, 2, 3, 3, 3]
a = 3, b = 2
Output : 1
Add each number once and multiply the sum by a, we will get a times the sum of each element of the array. Store it as a_sum. Subtract the sum of the whole array from the a_sum and divide the result by (a-b). The number we get is the required number (which appears b times in the array).
C++
// CPP program to find the only element that // appears b times #include <bits/stdc++.h> using namespace std; int appearsbTimes(int arr[], int n, int a, int b) { unordered_set<int> s; int a_sum = 0, sum = 0; for (int i = 0; i < n; i++) { if (s.find(arr[i]) == s.end()) { s.insert(arr[i]); a_sum += arr[i]; } sum += arr[i]; } a_sum = a * a_sum; return ((a_sum - sum) / (a - b)); } int main() { int arr[] = { 1, 1, 2, 2, 2, 3, 3, 3 }; int a = 3, b = 2; int n = sizeof(arr) / sizeof(arr[0]); cout << appearsbTimes(arr, n, a, b); return 0; } |
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Java
// Java program to find the only element that // appears b times import java.util.*; class GFG { static int appearsbTimes(int arr[], int n, int a, int b) { HashSet<Integer> s = new HashSet<Integer>(); int a_sum = 0, sum = 0; for (int i = 0; i < n; i++) { if (!s.contains(arr[i])) { s.add(arr[i]); a_sum += arr[i]; } sum += arr[i]; } a_sum = a * a_sum; return ((a_sum - sum) / (a - b)); } // Driver Code public static void main(String[] args) { int arr[] = { 1, 1, 2, 2, 2, 3, 3, 3 }; int a = 3, b = 2; int n = arr.length; System.out.println(appearsbTimes(arr, n, a, b)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 program to find the only # element that appears b times def appearsbTimes(arr, n, a, b): s = dict() a_Sum = 0 Sum = 0 for i in range(n): if (arr[i] not in s.keys()): s[arr[i]] = 1 a_Sum += arr[i] Sum += arr[i] a_Sum = a * a_Sum return ((a_Sum - Sum) // (a - b)) # Driver code arr = [1, 1, 2, 2, 2, 3, 3, 3] a, b = 3, 2n = len(arr) print(appearsbTimes(arr, n, a, b)) # This code is contributed by mohit kumar |
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C#
// C# program to find the only element that // appears b times using System; using System.Collections.Generic; class GFG { static int appearsbTimes(int []arr, int n, int a, int b) { HashSet<int> s = new HashSet<int>(); int a_sum = 0, sum = 0; for (int i = 0; i < n; i++) { if (!s.Contains(arr[i])) { s.Add(arr[i]); a_sum += arr[i]; } sum += arr[i]; } a_sum = a * a_sum; return ((a_sum - sum) / (a - b)); } // Driver Code public static void Main(String[] args) { int []arr = { 1, 1, 2, 2, 2, 3, 3, 3 }; int a = 3, b = 2; int n = arr.Length; Console.WriteLine(appearsbTimes(arr, n, a, b)); } } // This code is contributed by Princi Singh |
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Output:
1
Please refer below article for more approaches.
Find the only element that appears k times.
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