Common Child
Problem Statement :
A string is said to be a child of a another string if it can be formed by deleting 0 or more characters from the other string. Given two strings of equal length, what's the longest string that can be constructed such that it is a child of both? For example, ABCD and ABDC have two children with maximum length 3, ABC and ABD. They can be formed by eliminating either the D or C from both strings. Note that we will not consider ABCD as a common child because we can't rearrange characters and ABCD ABDC. Function Description Complete the commonChild function in the editor below. It should return the longest string which is a common child of the input strings. commonChild has the following parameter(s): s1, s2: two equal length strings Input Format There is one line with two space-separated strings, s1 and s2. Constraints 1 <= | s1 | , | s2 | <= 5000 All characters are upper case in the range ascii[A-Z]. Output Format Print the length of the longest string s, such that is a child of both s1 and s2.
Solution :
Solution in C :
In C++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
int dp[5005][5005];
int main() {
string a, b;
cin >> a >> b;
for(int i = 0; i < a.size(); ++i)
for(int j = 0; j < b.size(); ++j)
if(a[i] == b[j])
dp[i + 1][j + 1] = dp[i][j] + 1;
else
dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
cout << dp[a.size()][b.size()];
return 0;
}
In Java ;
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String s1 = s.nextLine();
String s2 = s.nextLine();
int[][] lengths = new int[s1.length()+1][s2.length()+1];
for (int i = 1; i < lengths.length; i++) {
for (int j = 1; j < lengths.length; j++) {
if (s1.charAt(i-1) == s2.charAt(j-1)) {
lengths[i][j] = lengths[i-1][j-1] + 1;
}
else if (lengths[i][j-1] > lengths[i-1][j]) {
lengths[i][j] = lengths[i][j-1];
}
else {
lengths[i][j] = lengths[i-1][j];
}
}
}
System.out.println(lengths[lengths.length-1][lengths.length-1]);
}
}
In C :
#include<stdio.h>
#include<string.h>
char str1[5005],str2[5005];
int c[5005][5005];
int main()
{
int i,j,m,n;
scanf("%s %s",str1+1,str2+1);
m=strlen(str1+1);
n=strlen(str2+1);
for(i=1;i<=m;i++)c[i][0]=0;
for(j=0;j<=n;j++)c[0][j]=0;
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
{
if(str1[i]==str2[j])
c[i][j]=c[i-1][j-1]+1;
else if(c[i-1][j]>=c[i][j-1])
c[i][j]=c[i-1][j];
else
c[i][j]=c[i][j-1];
}
printf("%d\n",c[m][n]);
return 0;
}
In Python3 :
def llis(a, b):
m = len(a)
n = len(b)
prev = [0 for x in range(n + 1)]
for i in range(1, m + 1):
curr = [0 for x in range(n + 1)]
for j in range(1, n + 1):
curr[j] = max(prev[j], curr[j - 1])
if a[i - 1] == b[j - 1]:
curr[j] = max(curr[j], prev[j - 1] + 1)
prev = curr
return curr[n]
if __name__ == '__main__':
a = input().strip()
b = input().strip()
c = set(a) & set(b)
a = "".join([x for x in a if x in c])
b = "".join([x for x in b if x in c])
print(llis(a, b))
View More Similar Problems
Jenny's Subtrees
Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .
View Solution →Tree Coordinates
We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For
View Solution →Array Pairs
Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .
View Solution →Self Balancing Tree
An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ
View Solution →Array and simple queries
Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →