Added codes for 29 and 30 May

GeeksForGeeks/May/29-5-24/GFG.java

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+//{ Driver Code Starts
+import java.io.*;
+import java.util.*;
+
+class GFG {
+    public static void main(String[] args) throws IOException {
+        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+        int t;
+        t = Integer.parseInt(br.readLine());
+        while (t-- > 0) {
+
+            int n;
+            n = Integer.parseInt(br.readLine());
+
+            int x;
+            x = Integer.parseInt(br.readLine());
+
+            int y;
+            y = Integer.parseInt(br.readLine());
+
+            Solution obj = new Solution();
+            int res = obj.findWinner(n, x, y);
+
+            System.out.println(res);
+        }
+    }
+}
+
+// } Driver Code Ends
+
+
+
+class Solution {
+    public static int findWinner(int n, int x, int y) {
+        // code here
+        int []dp= new int[n+1];
+        dp[1]=1;
+        for(int i=2;i<dp.length;i++)
+        {
+            if(dp[i-1]==0)
+                dp[i]=1;
+            else if(i-x>=0 && dp[i-x]==0)
+                dp[i]=1;
+            else if(i-y>=0 && dp[i-y]==0)
+                dp[i]=1;
+            else
+                dp[i]=0;
+        }
+
+        return dp[n]==1?1:0;
+    }
+}

GeeksForGeeks/May/29-5-24/README.md

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+Time complexity - O(n)
+Space complexity - O(n)

GeeksForGeeks/May/30-5-24/GFG.java

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+//{ Driver Code Starts
+import java.io.*;
+import java.util.*;
+
+class GFG {
+    public static void main(String[] args) throws IOException {
+        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+        int t;
+        t = Integer.parseInt(br.readLine());
+        while (t-- > 0) {
+
+            String s1;
+            s1 = br.readLine();
+
+            String s2;
+            s2 = br.readLine();
+
+            Solution obj = new Solution();
+            int res = obj.countWays(s1, s2);
+
+            System.out.println(res);
+        }
+    }
+}
+
+// } Driver Code Ends
+
+
+
+class Solution {
+    public static int countWays(String s1, String s2) {
+        // code here
+        int MOD = 1000000007;
+        int n = s1.length();
+        int m = s2.length();
+        int[][] dp = new int[n + 1][m + 1];
+
+        for (int i = 0; i <= n; i++)
+            dp[i][0] = 1;
+
+        for (int i = 1; i <= n; i++) 
+        {
+            for (int j = 1; j <= m; j++) 
+            {
+                if (s1.charAt(i - 1) == s2.charAt(j - 1))
+                {
+                    dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % MOD;
+                }
+                else
+                {
+                    dp[i][j] = dp[i - 1][j] % MOD;
+                }
+            }
+        }
+
+        return dp[n][m];
+    }
+}

GeeksForGeeks/May/30-5-24/README.md

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+Time complexity - O(n*m)
+Space complexity - O(n*m)

LeetCode/May/29-5-24/README.md

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+Time complexity - O(n)
+Space complexity - O(n)

LeetCode/May/29-5-24/Solution.java

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+class Solution 
+{
+    public int numSteps(String s) 
+    {
+        int ans = 0;
+        StringBuilder sb = new StringBuilder(s);
+
+        while (sb.charAt(sb.length() - 1) == '0') 
+        {
+            sb.deleteCharAt(sb.length() - 1);
+            ++ans;
+        }
+
+        if (sb.toString().equals("1"))
+            return ans;
+
+        // `s` is now odd, so add 1 to `s` and cost 1 step.
+        ++ans;
+
+        // All the 1s will become 0s and can be popped by 1 step.
+        // All the 0s will become 1s and can be popped by 2 steps (adding 1 then dividing by 2).
+        for (char c : sb.toString().toCharArray())
+            ans += c == '1' ? 1 : 2;
+
+        return ans;
+    }
+}

LeetCode/May/30-5-24/README.md

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+Time complexity - O(n^2)
+Space complexity - O(1)

LeetCode/May/30-5-24/Solution.java

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+class Solution 
+{
+    public int countTriplets(int[] arr) 
+    {
+        int count = 0;
+
+        for(int i = 0; i < arr.length; i++)
+        {
+            int xor = 0;
+            for(int j = i; j < arr.length; j++)
+            {
+                xor ^= arr[j];
+                if(xor == 0)
+                    count += (j - i);
+            }
+        }
+
+        return count;
+    }
+}