13November

3Sum Problem:

class Solution {
public:
    

vector<vector<int>> threeSum(vector<int>& nums) {
    vector<vector<int>> ans;
    int n = nums.size();
    
    // Sort the input array
    sort(nums.begin(), nums.end());
    
    // Loop through each element
    for (int i = 0; i < n; i++) {
        // Skip duplicate elements
        if (i > 0 && nums[i] == nums[i - 1]) {
            continue;
        }
        
        // Two pointers initialization
        int j = i + 1;
        int k = n - 1;
        
        // Two-pointer approach to find valid triplets
        while (j < k) {
            int sum = nums[i] + nums[j] + nums[k];
            
            if (sum < 0) {
                j++;  // Move the left pointer to increase the sum
            } 
            else if (sum > 0) {
                k--;  // Move the right pointer to decrease the sum
            } 
            else {
                // Found a triplet
                ans.push_back({nums[i], nums[j], nums[k]});
                
                
                j++;
                k--;
                
                
                while (j < k && nums[j] == nums[j - 1]) {
                    j++;
                }
                
                
                while (j < k && nums[k] == nums[k + 1]) {
                    k--;
                }
            }
        }
    }
    
    return ans;
}

};





set-mismatch:


class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        vector<int> ans;
        int n = nums.size();
        vector<bool> isPresent(n + 1, false);

        for (int i = 0; i < n; i++) {
            if (isPresent[nums[i]]) {
                ans.push_back(nums[i]);
            }
            isPresent[nums[i]] = true;
        }

        for (int i = 1; i < isPresent.size(); i++) {
            if (!isPresent[i]) {
                ans.push_back(i);
                break;
            }
        }
        return ans;
    }
};