Practice 11-Jun-2020

Author: sureshmangs

competitive programming/leetcode/2020-June-Challenge/Day-10-Search Insert Position.cpp

@@ -0,0 +1,40 @@
+Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You may assume no duplicates in the array.
+
+Example 1:
+
+Input: [1,3,5,6], 5
+Output: 2
+Example 2:
+
+Input: [1,3,5,6], 2
+Output: 1
+Example 3:
+
+Input: [1,3,5,6], 7
+Output: 4
+Example 4:
+
+Input: [1,3,5,6], 0
+Output: 0
+
+
+
+
+
+
+
+class Solution {
+public:
+    int searchInsert(vector<int>& nums, int target) {
+       int s=0, e=nums.size()-1, mid;
+        while(s<=e){
+            mid=s+(e-s)/2;
+            if(nums[mid]==target) return mid;
+            else if(nums[mid]>target) e=mid-1;
+            else s=mid+1;
+        }
+        return s;
+    }
+};

competitive programming/leetcode/35. Search Insert Position.cpp

@@ -0,0 +1,39 @@
+Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You may assume no duplicates in the array.
+
+Example 1:
+
+Input: [1,3,5,6], 5
+Output: 2
+Example 2:
+
+Input: [1,3,5,6], 2
+Output: 1
+Example 3:
+
+Input: [1,3,5,6], 7
+Output: 4
+Example 4:
+
+Input: [1,3,5,6], 0
+Output: 0
+
+
+
+
+
+
+class Solution {
+public:
+    int searchInsert(vector<int>& nums, int target) {
+       int s=0, e=nums.size()-1, mid;
+        while(s<=e){
+            mid=s+(e-s)/2;
+            if(nums[mid]==target) return mid;
+            else if(nums[mid]>target) e=mid-1;
+            else s=mid+1;
+        }
+        return s;
+    }
+};

dynamic programming/longest_common_subsequence_tabulation.cpp

@@ -0,0 +1,74 @@
+/*
+Given two sequences, find the length of longest subsequence present in both of them. Both the strings are of uppercase.
+
+Input:
+First line of the input contains no of test cases  T,the T test cases follow.
+Each test case consist of 2 space separated integers A and B denoting the size of string str1 and str2 respectively
+The next two lines contains the 2 string str1 and str2 .
+
+Output:
+For each test case print the length of longest  common subsequence of the two strings .
+
+Constraints:
+1<=T<=200
+1<=size(str1),size(str2)<=100
+
+Example:
+Input:
+2
+6 6
+ABCDGH
+AEDFHR
+3 2
+ABC
+AC
+
+Output:
+3
+2
+
+Explanation
+LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.
+
+LCS of "ABC" and "AC" is "AC" of length 2
+*/
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int LCS(string s1, string s2, int n, int m){
+    int dp[n+1][m+1];
+
+    for(int i=0;i<n+1;i++) dp[i][0]=0;
+    for(int j=0;j<m+1;j++) dp[0][j]=0;
+
+    for(int i=1;i<n+1;i++){
+        for(int j=1;j<m+1;j++){
+            if(s1[i-1]==s2[j-1])
+                 dp[i][j]=1 + dp[i-1][j-1];
+            else dp[i][j]=max(dp[i][j-1], dp[i-1][j]);
+        }
+    }
+    return dp[n][m];
+}
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    cout.tie(NULL);
+
+    int t;
+    cin>>t;
+    while(t--){
+        int n,m;
+        cin>>n>>m;
+        string s1,s2;
+        cin>>s1>>s2;
+        cout<<LCS(s1,s2,n,m)<<endl;
+    }
+
+return 0;
+}

dynamic programming/longest_common_substring_tabulation.cpp

@@ -0,0 +1,72 @@
+/*
+Given two strings X and Y. The task is to find the length of the longest common substring.
+
+Input:
+First line of the input contains number of test cases T. Each test case consist of three lines, first of which contains 2 space separated integers N and M denoting the size of string X and Y strings respectively. The next two lines contains the string X and Y.
+
+Output:
+For each test case print the length of longest  common substring of the two strings .
+
+Constraints:
+1 <= T <= 200
+1 <= N, M <= 100
+
+Example:
+Input:
+2
+6 6
+ABCDGH
+ACDGHR
+3 2
+ABC
+AC
+
+Output:
+4
+1
+
+Example:
+Testcase 1: CDGH is the longest substring present in both of the strings.
+*/
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int LCSubstring(string x, string y, int m, int n){
+    int dp[m+1][n+1];
+
+    for(int i=0;i<m+1;i++) dp[i][0]=0;
+    for(int j=0;j<n+1;j++) dp[0][j]=0;
+
+    int maxi=0;
+
+    for(int i=1;i<m+1;i++){
+        for(int j=1;j<n+1;j++){
+            if(x[i-1]==y[j-1])
+                dp[i][j]=1+dp[i-1][j-1];
+            else dp[i][j]=0;
+            maxi=max(maxi, dp[i][j]);
+        }
+    }
+    return maxi;
+}
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    cout.tie(NULL);
+    int t;
+    cin>>t;
+    while(t--){
+        int m,n;
+        cin>>m>>n;
+        string x, y;
+        cin>>x>>y;
+        cout<<LCSubstring(x,y,m,n)<<endl;
+    }
+
+return 0;
+}

dynamic programming/longest_palindromic_subsequence.cpp

@@ -0,0 +1,63 @@
+/*
+Given a String, find the longest palindromic subsequence
+
+Input:
+The first line of input contains an integer T, denoting no of test cases. The only line of each test case consists of a string S(only lowercase)
+
+Output:
+Print the Maximum length possible for palindromic subsequence.
+
+Constraints:
+1<=T<=100
+1<=|Length of String|<=1000
+
+
+Examples:
+Input:
+2
+bbabcbcab
+abbaab
+Output:
+7
+4
+*/
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int LCS(string s1, string s2, int n, int m){
+    int dp[n+1][m+1];
+
+    for(int i=0;i<n+1;i++) dp[i][0]=0;
+    for(int j=0;j<m+1;j++) dp[0][j]=0;
+
+    for(int i=1;i<n+1;i++){
+        for(int j=1;j<m+1;j++){
+            if(s1[i-1]==s2[j-1])
+                 dp[i][j]=1 + dp[i-1][j-1];
+            else dp[i][j]=max(dp[i][j-1], dp[i-1][j]);
+        }
+    }
+    return dp[n][m];
+}
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    cout.tie(NULL);
+
+    int t;
+    cin>>t;
+    while(t--){
+        string s1,s2;
+        cin>>s1;
+        int n=s1.length();
+        for(int i=n-1;i>=0;i--)  s2.push_back(s1[i]);
+        cout<<LCS(s1,s2,n,n)<<endl;
+    }
+
+return 0;
+}

dynamic programming/min_num_of_insertion_deletion.cpp

@@ -0,0 +1,75 @@
+/*
+Given two strings ‘str1’ and ‘str2’ of size m and n respectively. The task is to remove/delete and insert minimum number of characters from/in str1 so as to transform it into str2. It could be possible that the same character needs to be removed/deleted from one point of str1 and inserted to some another point.
+
+Input:
+
+The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains integers N and M denoting the length of the strings str1 and str2. Both the strings consist of only small letter alphabets.
+
+The second line of each test case contains the strings str1 and str2.
+
+
+Output:
+
+Print the total no of insertions and deletions to be done to convert str1 to str2. Output the answer in a new line.
+
+
+Constraints:
+
+1<= T <=100
+
+1<= N, M <=1000
+
+
+Example:
+
+Input:
+
+1
+
+4 3
+
+heap pea
+
+Output:
+
+3
+*/
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int LCS(string s1, string s2, int n, int m){
+    int dp[n+1][m+1];
+
+    for(int i=0;i<n+1;i++) dp[i][0]=0;
+    for(int j=0;j<m+1;j++) dp[0][j]=0;
+
+    for(int i=1;i<n+1;i++){
+        for(int j=1;j<m+1;j++){
+            if(s1[i-1]==s2[j-1])
+                 dp[i][j]=1 + dp[i-1][j-1];
+            else dp[i][j]=max(dp[i][j-1], dp[i-1][j]);
+        }
+    }
+    return dp[n][m];
+}
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    cout.tie(NULL);
+
+    int t;
+    cin>>t;
+    while(t--){
+        int n,m;
+        cin>>n>>m;
+        string s1,s2;
+        cin>>s1>>s2;
+        cout<<n+m-2*LCS(s1,s2,n,m)<<endl;
+    }
+
+return 0;
+}