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This approach involves sorting the input data first and then finding the solution by traversing through the sorted data. This approach is generally straightforward and often leads to a solution by leveraging sorted order, which simplifies many problems, such as finding pairs or detecting duplicates.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) or O(n), depending on the usage of additional data structures.
1import java.util.Arrays;
2
3public class Solution {
4 public static void solve(int[] arr) {
5 Arrays.sort(arr);
6 for (int num : arr) {
7 System.out.print(num + " ");
8 }
9 System.out.println();
10 }
11}This Java solution sorts the array using Arrays.sort and prints the elements. The problem-specific logic can be incorporated after sorting is complete.
This approach leverages a hash map to efficiently solve problems requiring quick lookups or to detect duplicates. This method is optimal for problems where you need to count occurrences or require O(1) average-time complexity for lookups.
Time Complexity: O(n)
Space Complexity: O(n) for the hash map.
1#include <unordered_map>
#include <vector>
void solve(std::vector<int>& arr) {
std::unordered_map<int, int> hash;
for(int num : arr) {
hash[num]++;
}
for(const auto& entry : hash) {
std::cout << entry.first << " appears " << entry.second << " times\n";
}
}This C++ solution uses an unordered map to store and count the occurrences of each number, providing efficient insertions and lookups.