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 <iostream>
2#include <unordered_map>
3using namespace std;
4
5void countElements(int arr[], int size) {
6 unordered_map<int, int> frequency;
7 for (int i = 0; i < size; i++) {
8 frequency[arr[i]]++;
9 }
10 for (auto& kvp : frequency) {
11 cout << "Element " << kvp.first << " appears " << kvp.second << " times\n";
12 }
13}
14
15int main() {
16 int arr[] = {1, 3, 1, 2, 3, 4, 5, 3};
17 int size = sizeof(arr) / sizeof(arr[0]);
18 countElements(arr, size);
19 return 0;
20}This C++ solution utilizes the unordered_map from the STL for efficient insertion and lookup operations which can be achieved in average O(1) time. It simplifies the implementation compared to managing a manual hash table.
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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int compare(const void* a, const void* b) {
5 return (*(int*)a - *(int*)b);
6}
7
8void countSortedElements(int arr[], int size) {
9 qsort(arr, size, sizeof(int), compare);
10 int count = 1;
11 for (int i = 1; i < size; i++) {
12 if (arr[i] == arr[i - 1]) {
13 count++;
14 } else {
15 printf("Element %d appears %d times\n", arr[i - 1], count);
16 count = 1;
17 }
18 }
19 printf("Element %d appears %d times\n", arr[size - 1], count);
20}
21
22int main() {
23 int arr[] = {1, 3, 1, 2, 3, 4, 5, 3};
24 int size = sizeof(arr) / sizeof(arr[0]);
25 countSortedElements(arr, size);
26 return 0;
27}This C solution sorts the input array using quicksort and then counts consecutive elements to tally occurrences. Sorting ensures similar elements cluster together, thus simplifying frequency determination.