Skip to main content

String Comparison Implemetation Recursively C++

Problem Statment: Write a recursive function to compare two cstrings.  Similiar to the strcmp() function return an integer greater than, equal to, or less than zero, accordingly as the string pointed to by s1 is greater than, equal to, or less than the string pointed to by s2.
  • "-1" if the first character that does not match has a  lower value in s1 than in s2
  • "0"    the contents of both strings are equal
  •  "1"    the first character that does not match has a greater value in s1 than in s2 


int str_cmp(char const* s1, char const* s2){

  if(*s1=='\0' && *s2=='\0') //if both character are null character
    return 0;       //they must be equal   
  if(*s1>*s2)
    return 1;
  if(*s1<*s2)
    return -1;

  return str_cmp(s1+1,s2+1);
}

Comments

Popular posts from this blog

Binary Tree Paths

Source: LeetCode Solution: We do a preorder traversal add each node in the string followed by an arrow "->" if the current node is a leaf node (have no right or left child) then we add the string to the list, otherwise, we add an arrow and make a recursive call in a preorder fashion. The previous string is stored in stack and we can use it later when the call is returned to add another possible routes from root to the leaf nodes.  Complexity Analysis Time complexity : O ( n ) . n nodes need to be traversed ( n represents the number of nodes in a given tree) .   Solution: public List< String > binaryTreePaths( TreeNode root) { ArrayList< String > path = new ArrayList (); binaryPath(root, "" ,path); return path; } public static void binaryPath( TreeNode root, String s, List< String > path){ if (root == null ) return ; ...

Non-decreasing Array - LeetCode

Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n). Example 1: Input: [4,2,3] Output: True Explanation: You could modify the first 4 to 1 to get a non-decreasing array.   Example 2: Input: [4,2,1] Output: False Explanation: You can't get a non-decreasing array by modify at most one element. Note: The n belongs to [1, 10,000]. Source: LeetCode At first glance, it looks like an easy and straightforward problem, but believe me it's greedy problem. In the problem, you are allowed to make atmost "one" modification to make a non-decreasing array. Nondecreasing means that every element is greater than or equal to the one before it e.g. [1,3,4,4,5,6,9]. For example, [2,2,3,2,4] can this become non-decreasing? Yes, by replacing 3 with 2 =>[2,2,2,2,4]. What about this:...

Merge Two Binary Trees

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree. Example 1: Input: Tree 1 Tree 2 1 2 / \ / \ 3 2 1 3 / \ \ 5 4 7    Output: Merged tree: 3 / \ 4 5 / \ \ 5 4 7   Note: The merging process must start from the root nodes of both trees.  Source: LeetCode Solution: We traverse both trees in a p...