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Two Pointer Problems


Here we have given some important two pointer problems below:

1. Valid Palindrome:

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string s, return true if it is a palindrome, or false otherwise.

 -> Solution:

class Solution {
public:
    bool isPalindrome(string s) {
      int low=0;
      int high=s.length()-1;
      while(low<high){
          while(!iswalnum(s[low]) && low<high){
              low++;
          }
          while(!iswalnum(s[high]) && low<high){
              high--;
          }
          if(tolower(s[low])!=tolower(s[high])){
           return false;
          }
          low++;
          high--;
      }
        return true;
    }
};

Time complexity: O(n)


2. Two Sum II:

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order,find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.
-> Solution:
class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        vector<int>vec;
        int st=0;
        int en=numbers.size()-1;
        while(st<en){
            int sum = numbers[st]+numbers[en];
            if(sum<target){
                st++;
            }
            else if(sum>target){
                en--;
            }
            else{
                vec.push_back(st+1);
                vec.push_back(en+1);
                break;
            }
        }
        return vec;
    }
};

Time complexity: O(n)


3. 3SUM:

Given an integer array nums, return all the triplets [nums[i], nums[j],

nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>>vec;
        for(int i=0;i<nums.size();i++){
            int ans = nums[i];
            if(i>0 && nums[i-1]==ans) continue;
            else{
                int st=i+1,en=nums.size()-1;
                while(st<en){
                    if(ans+nums[st]+nums[en]>0) en--;
                    else if (ans+nums[st]+nums[en]<0) st++;
                    else{
                        vec.push_back({ans,nums[st],nums[en]});
                    while(st<en && nums[st]==nums[st+1]){
                        st++;
                    }
                    st++;
                    while(st<en && nums[en]==nums[en-1]){
                        en--;
                    }
                    en--;
                    }
                }
            }
        }
        return vec;
    }
};

Time complexity: O(n^2)

Problem link: 3Sum - LeetCode
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