This approach involves using a HashMap (or dictionary in Python) to count the occurrences of each element in the array. This will help in efficiently checking if an element exists and how many times it appears, which is useful for problems requiring frequency analysis.
Time Complexity: O(n), as it iterates over the array once. Space Complexity: O(1) if the element range is known and constant, or O(k) where k is the range of elements.
1#include <stdio.h>
2#include <stdlib.h>
3
4void countElements(int arr[], int size) {
5 int* hashTable = (int*)calloc(1000, sizeof(int)); // Assuming elements range from 0 to 999
6 for (int i = 0; i < size; i++) {
7 hashTable[arr[i]]++;
8 }
9 for (int i = 0; i < 1000; i++) {
10 if (hashTable[i] > 0) {
11 printf("Element %d appears %d times\n", i, hashTable[i]);
12 }
13 }
14 free(hashTable);
15}
16
17int main() {
18 int arr[] = {1, 3, 1, 2, 3, 4, 5, 3};
19 int size = sizeof(arr) / sizeof(arr[0]);
20 countElements(arr, size);
21 return 0;
22}This C solution uses an array to simulate a hash table where each index represents an element and its value represents the frequency. This allows both insertion and lookup to occur in constant O(1) time.
Another approach is sorting the array and then identifying repeated elements by comparison with neighboring elements. This leverages the efficiency of modern sorting algorithms to bring repeated elements together, making it trivial to count consecutive duplicates.
Time Complexity: O(n log n) due to the sort operation. Space Complexity: O(1) if in-place sorting is considered.
1def count_sorted_elements(arr):
2 arr.sort()
3 count = 1
4 for i in range(1, len(arr)):
5 if arr[i] == arr[i - 1]:
6 count += 1
7 else:
8 print(f'Element {arr[i - 1]} appears {count} times')
9 count = 1
10 print(f'Element {arr[-1]} appears {count} times')
11
12arr = [1, 3, 1, 2, 3, 4, 5, 3]
13count_sorted_elements(arr)This Python method sorts the array and iterates through it to count appearances of each number. The sorted array ensures that duplicates are counted consecutively.