Find k closest elements in a sorted array using Java ......

here is the code .....


/*Given a sorted array arr[] and a value X, find the k closest elements to X in arr[].
 Examples:

 Input: K = 4, X = 35
 arr[] = {12, 16, 22, 30, 35, 39, 42,
 45, 48, 50, 53, 55, 56}
 Output: 30 39 42 45
 Note that if the element is present in array, then it should not be in output, only the other closest elements are required.
 */

       

public class FindKClosestElements {

 public static void main(String args[]) {

  int[] arr = {
   12,
   16,
   22,
   30,
   35,
   39,
   42,
   45,
   48,
   50,
   53,
   55,
   56
  };
  int element = 100;
  int k = 5;
  int m;
  int n;
  int distance1;
  int distance2;

  // case if element is is larger than even the last element in the array
  if (arr[arr.length - 1] < element) {
   n = arr.length - 1;
   while (k != 0) {
    System.out.print(arr[n]);
    System.out.print(" ");
    k--;
    n--;
   }
   return;
  }

  for (int i = 0; i < arr.length; i++) {
   if (arr[i] >= element) {
    m = i - 1;
    if (arr[i] == element)
     n = i + 1;
    else
     n = i;
    while (m >= 0 && n < arr.length && k != 0) {

     distance1 = arr[n] - element;
     distance2 = element - arr[m];

     if (distance1 <= distance2) {
      System.out.print(arr[n]);
      System.out.print(" ");
      k--;
      n++;
     } else {
      System.out.print(arr[m]);
      System.out.print(" ");
      k--;
      m--;
     }
    }
    if (k != 0) {
     if (m > 0)
      while (k != 0) {
       System.out.print(arr[m]);
       System.out.print(" ");
       m--;
       k--;
      } else
      while (k != 0) {
       System.out.print(arr[n]);
       System.out.print(" ");
       k--;
       n++;
      }
    }
    if (k == 0)
     return;
    else
     continue;
   }
  }
 }


}

       
 

Merging two sorted arrays using Java ....

here is the code .....

       
public class MergingSortedLists {

 public static void main(String args[]) {

  int[] arr1 = {
   1,
   4,
   5,
   43,
   56,
   59
  };
  int[] arr2 = {
   2,
   3,
   6,
   34,
   36,
   42,
   67
  };
  int[] sortedList = new int[arr1.length + arr2.length];
  mergeLists(arr1, arr2, sortedList);
 }

 public static void mergeLists(int[] arr1, int[] arr2, int[] sortedList) {
  int k = 0;
  int j = 0;
  int t = 0;
  while (k < arr1.length && j < arr2.length) {
   if (arr1[k] <= arr2[j]) {
    sortedList[t] = arr1[k];
    t++;
    k++;
   } else {
    sortedList[t] = arr2[j];
    t++;
    j++;
   }
  }
  if (k != arr1.length) { //case where arr1 is left with elements not added to sorted list
   while (k < arr1.length) {
    sortedList[t] = arr1[k];
    t++;
    k++;
   }
  }
  if (j != arr2.length) { //case where arr2 is left with elements not added to sorted list
   while (j < arr2.length) {
    sortedList[t] = arr2[j];
    t++;
    j++;
   }
  }
  for (int s: sortedList) {
   System.out.print(s);
   System.out.print(" ");

  }
 }
}

       
 

Check whether two numbers are of opposite signs or not using Java .....

Problem -Two numbers are given ,you have to evaluate whether they are of opposite signs or not

Logic -Numbers are represented in 2's complemented form in computer where sign bit 0 means it is 

            positive number and 1 means it is negative .

here is the code ......

       
public class CheckOppositeSigns {

 public static void main(String args[]) {
  int x = 2;
  int y = -13;
  if (checkForOppositeSigns(x, y))
   System.out.print("Yes both the integers are of opposite signs");
  else
   System.out.print("Both the integers are of same sign");
 }

 public static boolean checkForOppositeSigns(int x, int y) {

  int signofFirstVariable = (x >> 31) ^ (0x1); //shifting the variable to place the sign bit at unit                                                                                     //position and then XOR with 000000000
  int signofSecondVariable = (y >> 31) ^ (0x1);
  if (signofFirstVariable != signofSecondVariable)
   return true;

  return false;
 }
}