Here is a detailed Markdown blog post for the "Pairs of equals Sum" problem:

Pairs of equals Sum

Summary

Given an array nums of integers, find all pairs (i, j) such that nums[i] + nums[j] = 4006. This problem involves using arrays and hash data structures to solve a simple sum-equals problem.

Detailed Explanation

To solve this problem, we will use a two-pointer technique. We first sort the array in ascending order. Then, we initialize two pointers, one at the start of the array (i) and one at the end of the array (j). We calculate the sum nums[i] + nums[j] and compare it to our target sum 4006. If the sum is less than 4006, we increment i to increase the sum. If the sum is greater than 4006, we decrement j to decrease the sum.

Here's a step-by-step breakdown of the solution:

1. Sort the array in ascending order.
2. Initialize two pointers, i and j, at the start and end of the array respectively.
3. Calculate the sum nums[i] + nums[j].
4. Compare the sum to our target sum 4006. If the sum is less than 4006, increment i to increase the sum.
5. If the sum is greater than 4006, decrement j to decrease the sum.
6. Repeat steps 3-5 until i and j meet or cross each other.

The time complexity of this solution is O(n log n) due to the sorting step, where n is the size of the array. The space complexity is O(1) as we only use a few extra variables to store the pointers and sum.

Optimized Solutions

Java

java
import java.util.Arrays;

public class PairsOfEqualsSum {
    public static void findPairs(int[] nums, int target) {
        Arrays.sort(nums);
        int i = 0, j = nums.length - 1;
        while (i < j) {
            int sum = nums[i] + nums[j];
            if (sum == target) {
                System.out.println("Pair found: (" + i + ", " + j + ")");
                i++;
                j--;
            } else if (sum < target) {
                i++;
            } else {
                j--;
            }
        }
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4, 5};
        findPairs(nums, 4006);
    }
}

Python

python
def find_pairs(nums, target):
    nums.sort()
    i, j = 0, len(nums) - 1
    while i < j:
        sum = nums[i] + nums[j]
        if sum == target:
            print("Pair found: ({}, {})".format(i, j))
            i += 1
            j -= 1
        elif sum < target:
            i += 1
        else:
            j -= 1

nums = [1, 2, 3, 4, 5]
find_pairs(nums, 4006)

C++

cpp
#include <iostream>
#include <vector>
#include <algorithm>

void findPairs(std::vector<int>& nums, int target) {
    std::sort(nums.begin(), nums.end());
    int i = 0, j = nums.size() - 1;
    while (i < j) {
        int sum = nums[i] + nums[j];
        if (sum == target) {
            std::cout << "Pair found: (" << i << ", " << j << ")" << std::endl;
            i++;
            j--;
        } else if (sum < target) {
            i++;
        } else {
            j--;
        }
    }
}

int main() {
    std::vector<int> nums = {1, 2, 3, 4, 5};
    findPairs(nums, 4006);
    return 0;
}

JavaScript

javascript
function findPairs(nums, target) {
    nums.sort((a, b) => a - b);
    let i = 0, j = nums.length - 1;
    while (i < j) {
        let sum = nums[i] + nums[j];
        if (sum === target) {
            console.log("Pair found: (" + i + ", " + j + ")");
            i++;
            j--;
        } else if (sum < target) {
            i++;
        } else {
            j--;
        }
    }
}

let nums = [1, 2, 3, 4, 5];
findPairs(nums, 4006);

These optimized solutions use the same two-pointer technique as described in the detailed explanation section.