main changes

Author: HarshitChari

Backtracking/palindrome_partition.py

@@ -0,0 +1,30 @@
+# Python code for implementation of Naive Recursive
+# approach
+def isPalindrome(x):
+return x == x[::-1]
+
+
+def minPalPartion(string, i, j):
+if i >= j or isPalindrome(string[i:j + 1]):
+return 0
+ans = float('inf')
+for k in range(i, j):
+count = (
+1 + minPalPartion(string, i, k)
++ minPalPartion(string, k + 1, j)
+)
+ans = min(ans, count)
+return ans
+
+
+def main():
+string = "ababbbabbababa"
+print(
+"Min cuts needed for Palindrome Partitioning is ",
+minPalPartion(string, 0, len(string) - 1),
+)
+
+if __name__ == "__main__":
+main()
+
+# This code is contributed by itsvinayak

Leetcode - Top Interview Questions/3sum.cpp

@@ -1,34 +1,40 @@
-// Time: O(n^2)
-// Space: O(1)
-
 class Solution {
 public:
- vector<vector<int>> threeSum(vector<int> &nums) {
- sort(begin(nums), end(nums));
- vector<vector<int>> result;
- for (int i = size(nums) - 1; i >= 2; --i) {
- if (i + 1 < size(nums) && nums[i] == nums[i + 1]) {
+ vector<vector<int>> threeSum(vector<int>& nums) {
+ vector<vector<int>>ans;
+ vector<int> temp;
+ sort(nums.begin(), nums.end());
+ if(nums.size()<3)
+ return ans;
+ int i,j,k;
+ for(i=0;i<nums.size()-2;i++){
+ if(i!=0 && nums[i]==nums[i-1])
  continue;
- }
- const auto& target = -nums[i];
- int left = 0, right = i - 1;
- while (left < right) {
- if (nums[left] + nums[right] < target) {
- ++left;
- } else if (nums[left] + nums[right] > target) {
- --right;
- } else {
- result.push_back({nums[left], nums[right], nums[i]});
- ++left; --right;
- while (left < right && nums[left] == nums[left - 1]) {
- ++left;
- }
- while (left < right && nums[right] == nums[right + 1]) {
- --right;
- }
+ j=i+1;
+ k=nums.size()-1;
+ 
+ while(j<k){
+ if(-nums[i]==nums[j]+nums[k]){
+ temp.clear();
+ temp.push_back(nums[i]);
+ temp.push_back(nums[j]);
+ temp.push_back(nums[k]);
+ // if(find(ans.begin(),ans.end(),temp)==ans.end()){
+ ans.push_back(temp);
+ //cout<<"in if";
+ // }
+ while(j<k && nums[j]==nums[j+1]) 
+ j++;
+ j++;
+ 
+ k--;
  }
+ else if(-nums[i]<nums[j]+nums[k])
+ k--;
+ else
+ j++;
  }
  }
- return result;
+ return ans; 
  }
-};
+};