SDE - 1

string longestPalindrome(string s) {
int n = s.size();
if(n==0) return "";
vector> dp(n, vector(n, false));
for(int i=0;i=0;i--) {
for(int j=i+1;j if(s[i] == s[j]) {
//j-i==1 means the string is of two characters and since both are equal it will be pallindrome
//dp[i+1][j-1] means the the string between ith and jth character is a pallindrome
//we know that adding same character at the left and right end of pallindrome creates another pallindrome
if(j-i==1 || dp[i+1][j-1]) {
dp[i][j] = true;
//if the length of new pallindrome is greater than previous pallindrome 
//we need to update the value of the string
if(ans.size() < j-i+1) {
ans = s.substr(i, j-i+1);
}
}
}
}
}
return ans;
}

Tip 1 : notmalization and real life examples of its forms

Tip 1 : read all types of joins and its examples.

Take three variables; let’s call them first, second and third and mark them as +∞.
Iterate through the array and for each element (let’s call it current), 
Check if first>current, assign the first value to the second and second value to third and assign current to first.
If the above step is not true then the current element might be a candidate of the second smallest element, so check if currentIf the above step is not true then the current element might be a candidate of the third-smallest element, so check if currentAt the end print the third, it will third-smallest element in the array

Tip 1 : prepare a small introduction 
Tip 2 : Dont hesitate in HR round . Be confident while answering