Merge pull request #637 from arpitghura/main

Refactorised Java Folder

Author: gantavyamalviya

Java/.idea/.gitignore

@@ -1,60 +1,60 @@
-package com.company;
-
-public class LeftRotateArrayByD {
-
-    //Reversal Algorithm-->
-    static void reverse(int[] arr, int low, int high){
-        int temp=0;
-        while (low<high) {
-            temp = arr[low];
-            arr[low] = arr[high];
-            arr[high] = temp;
-            low++;
-            high--;
-        }
-    }
-
-    static void leftRotate(int[] arr, int d){
-        if (d==0)
-            return;
-
-        int n=arr.length;
-        reverse(arr,0,d-1);
-        reverse(arr, d,n-1);
-        reverse(arr,0,n-1);
-    }
-
-    // Better Method-->
-
-//    static void leftRotate(int[] arr, int d){
-//        int[] temp = new int[d];
-//        for (int i=0; i<d; i++)
-//            temp[i]=arr[i];
-//        for (int i=d; i< arr.length; i++)
-//            arr[i-d]=arr[i];
-//        for (int i=0; i<d; i++)
-//            arr[arr.length-d+1]=temp[i];
-//    }
-
-    // Naive Method-->
-
-//    static void leftRotateOne(int[] arr){
-//        int n = arr.length;
-//        int temp = arr[0];
-//        for (int i=1; i<arr.length; i++)
-//            arr[i-1]=arr[i];
-//        arr[n-1]=temp;
-//    }
-//    static void LeftRotate(int[] arr, int d){
-//        for (int i=0; i<d; i++)
-//            leftRotateOne(arr);
-//    }
-    public static void main(String[] args) {
-        int[] arr = new int[]{1,2,3,4,5};
-        int d = 2;
-        int n = arr.length;
-        leftRotate(arr, 2);
-        for (int i=0; i<n; i++)
-            System.out.print(arr[i] + " ");
-    }
-}
+package com.company;
+
+public class LeftRotateArrayByD {
+
+    //Reversal Algorithm-->
+    static void reverse(int[] arr, int low, int high){
+        int temp=0;
+        while (low<high) {
+            temp = arr[low];
+            arr[low] = arr[high];
+            arr[high] = temp;
+            low++;
+            high--;
+        }
+    }
+
+    static void leftRotate(int[] arr, int d){
+        if (d==0)
+            return;
+
+        int n=arr.length;
+        reverse(arr,0,d-1);
+        reverse(arr, d,n-1);
+        reverse(arr,0,n-1);
+    }
+
+    // Better Method-->
+
+//    static void leftRotate(int[] arr, int d){
+//        int[] temp = new int[d];
+//        for (int i=0; i<d; i++)
+//            temp[i]=arr[i];
+//        for (int i=d; i< arr.length; i++)
+//            arr[i-d]=arr[i];
+//        for (int i=0; i<d; i++)
+//            arr[arr.length-d+1]=temp[i];
+//    }
+
+    // Naive Method-->
+
+//    static void leftRotateOne(int[] arr){
+//        int n = arr.length;
+//        int temp = arr[0];
+//        for (int i=1; i<arr.length; i++)
+//            arr[i-1]=arr[i];
+//        arr[n-1]=temp;
+//    }
+//    static void LeftRotate(int[] arr, int d){
+//        for (int i=0; i<d; i++)
+//            leftRotateOne(arr);
+//    }
+    public static void main(String[] args) {
+        int[] arr = new int[]{1,2,3,4,5};
+        int d = 2;
+        int n = arr.length;
+        leftRotate(arr, 2);
+        for (int i=0; i<n; i++)
+            System.out.print(arr[i] + " ");
+    }
+}

Java/FindGreatestCommonDivisorOfArray.java renamed to Java/Array/FindGreatestCommonDivisorOfArray.java

@@ -1,55 +1,55 @@
-package XtraQuestions.StringsArrayMixed;
-// You are given two numbers n and k. You are required to rotate n k times to the right. If kis positive, rotate to the right l.e. remove rightmost digit and make it leftmost. Do the reverse for negative value of k. Also k can have an absolute value larger than number of digits in n. 2. Take as input n and k. 3. Print the rotated number. 4. 
-// Note - Assume that the number of rotations will not cause leading O's in the result, e.g. such an input will not be given
-// n = 12340056 k= 3 r = 05612340
-// Constraints 1 <= n < 10^9 - 10^9 < k < 10^9
-// Format
-// Input
-// "n" where n is any integer. "K" where k is any integer.
-// Output
-// "r", the rotated number
-// SAMPLE INPUT
-// 562984
-// 2
-// OUTPUT
-// 845629
-
-import java.util.*;
-
-public class RoatateNumberInRightKTimes {
-    public static void main(String[] args) {
-        // int n = 12340056; // This type of number is not allowed
-        int n = 124452;
-        System.out.println(n);
-        
-        int k = 2;
-        
-        int frontnum = 0;
-        for(int i=0; i<k; i++){
-            int rem = n%10;
-            int remaining = n/10;
-            
-            // for 0's in between we have to use this
-            // if(rem==0){
-            //     String padded = String.format("%08d" , remaining);
-            //     // System.out.println("padded: "+ padded);
-            //     n = Integer.parseInt(padded);
-            //     // n = padded;
-            //     System.out.println("the n is : " + n);
-            //     continue;
-            // }
-            int count = 0;
-            int temp = remaining;
-            while(temp!=0){
-                count++;
-                temp = temp/10;
-            }
-            // System.out.println("count : " + count);
-            frontnum = rem*(int)(Math.pow(10,count)) + remaining;
-            n = frontnum;
-            // System.out.println("frontnum : " + frontnum);
-        }
-
-        System.out.println("The number after " + k + " rotations: " + frontnum);
-    }
-}
+package XtraQuestions.StringsArrayMixed;
+// You are given two numbers n and k. You are required to rotate n k times to the right. If kis positive, rotate to the right l.e. remove rightmost digit and make it leftmost. Do the reverse for negative value of k. Also k can have an absolute value larger than number of digits in n. 2. Take as input n and k. 3. Print the rotated number. 4. 
+// Note - Assume that the number of rotations will not cause leading O's in the result, e.g. such an input will not be given
+// n = 12340056 k= 3 r = 05612340
+// Constraints 1 <= n < 10^9 - 10^9 < k < 10^9
+// Format
+// Input
+// "n" where n is any integer. "K" where k is any integer.
+// Output
+// "r", the rotated number
+// SAMPLE INPUT
+// 562984
+// 2
+// OUTPUT
+// 845629
+
+import java.util.*;
+
+public class RoatateNumberInRightKTimes {
+    public static void main(String[] args) {
+        // int n = 12340056; // This type of number is not allowed
+        int n = 124452;
+        System.out.println(n);
+        
+        int k = 2;
+        
+        int frontnum = 0;
+        for(int i=0; i<k; i++){
+            int rem = n%10;
+            int remaining = n/10;
+            
+            // for 0's in between we have to use this
+            // if(rem==0){
+            //     String padded = String.format("%08d" , remaining);
+            //     // System.out.println("padded: "+ padded);
+            //     n = Integer.parseInt(padded);
+            //     // n = padded;
+            //     System.out.println("the n is : " + n);
+            //     continue;
+            // }
+            int count = 0;
+            int temp = remaining;
+            while(temp!=0){
+                count++;
+                temp = temp/10;
+            }
+            // System.out.println("count : " + count);
+            frontnum = rem*(int)(Math.pow(10,count)) + remaining;
+            n = frontnum;
+            // System.out.println("frontnum : " + frontnum);
+        }
+
+        System.out.println("The number after " + k + " rotations: " + frontnum);
+    }
+}

Java/FindNoWithEvenNoOfDigits.java renamed to Java/Array/FindNoWithEvenNoOfDigits.java

@@ -1,43 +1,43 @@
-package Searching;
-
-
-public class BinarySearchSQRT {
-    public static void main(String[] args) {
-        int n = 40;
-        int p = 3;
-        System.out.printf("%.3f",sqrt(n,p));
-    
-    }
-    //time complexity: O(log n)
-    static double sqrt(int n, int p)
-    {
-        int start = 0;
-        int end = n;
-        double root = 0.0;
-
-        while(start <= end)
-        {
-            int m = start + (end - start) / 2;
-
-            if(m * m == n)
-            return m;
-            if(m * m > n)
-            {
-                end = m - 1;
-            }
-            else
-            start = m + 1;
-        }
-        double incr = 0.1;
-        for(int i = 0; i < p; i++)
-        {
-            while(root * root <= n)
-            root += incr;
-            root -= incr;
-            incr /= 10;
-
-        }
-        return root;
-        }
-    }
-
+package Searching;
+
+
+public class BinarySearchSQRT {
+    public static void main(String[] args) {
+        int n = 40;
+        int p = 3;
+        System.out.printf("%.3f",sqrt(n,p));
+    
+    }
+    //time complexity: O(log n)
+    static double sqrt(int n, int p)
+    {
+        int start = 0;
+        int end = n;
+        double root = 0.0;
+
+        while(start <= end)
+        {
+            int m = start + (end - start) / 2;
+
+            if(m * m == n)
+            return m;
+            if(m * m > n)
+            {
+                end = m - 1;
+            }
+            else
+            start = m + 1;
+        }
+        double incr = 0.1;
+        for(int i = 0; i < p; i++)
+        {
+            while(root * root <= n)
+            root += incr;
+            root -= incr;
+            incr /= 10;
+
+        }
+        return root;
+        }
+    }
+