Java Interview Question.

1. Check if array has duplicate elements or not?

//using for loop
public boolean checkDubli(int[] input) {

  for (int i = 0; i < input.length; i++) {
    for (int j = i + 1; j < input.length; j++) {
      if (input[i] == input[j]) {
        return true;
      }
    }
  }
  return false;
}

//using hash map
public boolean checkDublicate(int[] input) {
  Map<Integer, Integer> map = new HashMap<>();
  for (int i = 0; i < input.length; i++) {
    if (map.containsKey(input[i])) {
      return true;
    } else {
      map.put(input[i], 1);
    }
  }
  return false;
}

2. Give an algorithm for finding the element which appears maximum number of times in array.

//method 1.

public void findMAxFreqNum(int[] input) {
  int count;
  int maxCount = 0;
  int maxRepNum = 0;
  for (int j : input) {
    count = 0;
    for (int k : input) {
      if (j == k) {
        count++;
      }
    }
    if (count > maxCount) {
      maxCount = count;
      maxRepNum = j;
    }
  }
  if (maxCount == 1) {
    System.out.println("all numbers are unique");
  } else {
    System.out.println(maxRepNum + "");
  }

}
// method 2
public void findMAxFreqNum2(int[] arr) {
  Arrays.sort(arr);
  int count;
  int maxCount = 1;
  int maxFreqNum = arr[0];
  for (int i = 1; i < arr.length; i++) {
    count = 1;
    while (arr[i] == arr[i - 1]) {
      count++;
      i++;
    }
    if (count > maxCount) {
      maxCount = count;
      maxFreqNum = arr[i-1];
    }
  }
  if (maxCount == 1) {
    System.out.println("All numbers are unique");
  } else {
    System.out.println(maxFreqNum);
  }
}

3. In array find the first element in the array which is repeated.

//using map
public int firstElementRep1(int[] arr1) {
  Map<Integer, Integer> map = new HashMap<>();
  for (int i = 0; i < arr1.length; i++) {
    if (map.containsKey(arr1[i])) {
      return arr1[i];
    } else {
      map.put(arr1[i], 1);
    }
  }
  return 0;
}

4. Given an array of n element, find two elements in the array such that their sum is equal to given element K.

public void findElementSum(int[] arr, int k) {
  int sum = 0;
  for (int i = 0; i < arr.length; i++) {
    for (int j = i + 1; j < arr.length; j++) {
      sum = arr[i] + arr[j];
      if (sum == k) {
        System.out.println("two element are:" + arr[i] + " , " + arr[j] + " = " + k);
      }
    }
  }
}

5. Find three element in the arrya such that their sum is equal to given element K.

public void findThreeNumSum(int[] arr, int s) {
  int sum;
  for (int i = 0; i < arr.length; i++) {
    for (int k = i + 1; k < arr.length; k++) {
      for (int j = k + 1; j < arr.length; j++) {
        sum = arr[i] + arr[k] + arr[j];
        if (sum == s) {
          System.out.println(
              "Three elements are:" + arr[i] + " , " + arr[k] + " , " + arr[j] + " = " + s);
        }
      }

    }
  }
}

6.Given a sorted array of n integer that has been rotated an unknown number of times, give a O(logn) algorith that finds an element in the array.

// First learn Binary Search

public int searchNum(int[] arr, int k) {
  int mid;
  int l = 0, r = arr.length - 1;
  while (l <= r) {
    mid = l + ((r - l) / 2);
    if (k == arr[mid]) {
      return mid;
    } else if (k > arr[mid]) {
      l = mid + 1;
    } else if (k < arr[mid]) {
      r = mid - 1;
    }
  }
  return -1;
}

//Actual code is pending

7. Given a sorted array A of n element, possibly with duplicates. Find the index of the first occurrence of a number in O(logn) time.

public int findDupNumFirstPos(int[] arr, int num) {
  int l = 0, r = arr.length - 1;
  int mid;
  int searchItemPos = -1;
  while (l <= r) {
    mid = l + ((r - l) / 2);
    if (num == arr[mid]) {
      searchItemPos = mid;
      r = mid - 1;
    } else if (num > arr[mid]) {
      l = mid + 1;
    } else if (num < arr[mid]) {
      r = mid - 1;
    }
  }
  return searchItemPos;
}

8. Separate even and odd number in array. Put all even number first and then odd number.

public void sepEvenOdd(int[] arr) {
  int l = 0, r = arr.length - 1, temp;
  while (l < r) {
    if (arr[l] % 2 == 0) {
      l++;
    }
    while (arr[r] % 2 != 0) {
      r--;
    }
    if (l < r) {
      temp = arr[l];
      arr[l] = arr[r];
      arr[r] = temp;
    }
  }
  for (int k : arr) {
    System.out.print(k + " , ");
  }
}

9 Sort an array of 0,1 and 2.

// method 1 simple way.

public void sortZeroOneTwo(int[] arr) {
  Arrays.sort(arr);
  for (int i : arr) {
    System.out.print(i + " ");
  }
}

// using loop
public void sortZeroOneTwo2(int[] arr) {
  int count0 = 0, count1 = 0, count2 = 0;
  for (int i = 0; i < arr.length; i++) {
    if (arr[i] == 0) {
      count0++;
    } else if (arr[i] == 1) {
      count1++;
    } else if (arr[i] == 2) {
      count2++;
    }
  }
  for(int i=0;i<count0-1;i++){
    arr[i]=0;
  }
  for (int i=count0;i<count0+count1;i++){
    arr[i]=1;
  }
  for (int i=count0+count1;i<count0+count1+count2;i++){
    arr[i]=2;
  }
  System.out.println();
  for (int i:arr){
    System.out.print(i+" ");
  }
}