Day1 Arrays - My LeetCode Diary

int[] myNum = {10, 20, 30, 40};
//Loop Through an Array
for (int i = 0; i < myNum.length; i++) {
  System.out.println(myNum[i]);
}

//OR
for (int i : myNum) {
  System.out.println(i);
}

int[][] myNumbers = { {1, 2, 3, 4}, {5, 6, 7} };
System.out.println(myNumbers[1][2]); // Outputs 7
//Loop Through Multi-Dimensional Arrays
for (int i = 0; i < myNumbers.length; ++i) {
      for(int j = 0; j < myNumbers[i].length; ++j) {
        System.out.println(myNumbers[i][j]);
      }
}

Binary Search Notes

Prerequisites

  • Sorted Arrays
  • No Duplicates

Left close, right close[left, right]

  • while (left <= right) use <=
  • if (nums[middle] > target) right = middle - 1, because num[mid] will not equal to target number, the right index could be mid - 1.
  • O(log n)
class Solution {
    public int search(int[] nums, int target) {
        if (target < nums[0] || target > nums[nums.length - 1]) {
            return -1;
        }
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + ((right - left) >> 1);
            if (nums[mid] == target)
                return mid;
            else if (nums[mid] < target)
                left = mid + 1;
            else if (nums[mid] > target)
                right = mid - 1;
        }
        return -1;
    }
}

Left close, right open[left,right)

class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = left + ((right - left) >> 1);
            if (nums[mid] == target)
                return mid;
            else if (nums[mid] < target)
                left = mid + 1;
            else if (nums[mid] > target)
                right = mid;
        }
        return -1;
    }
}

Link

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

Constraints:

  • 1 <= nums.length <= 104
  • 104 < nums[i], target < 104
  • All the integers in nums are unique.
  • nums is sorted in ascending order.

My answers

class Solution {
    public int search(int[] nums, int target) {
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == target){
                return i;
            }
        }
        return -1;
    }
}

//[left,right)
class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length;
        while(left < right) {
            int mid = left + ((right - left) >> 1);
            if(target == nums[mid]){
                return mid;
            } else if(nums[mid] < target) {
                left = mid + 1;
            } else if(nums[mid] > target) {
                right = mid;
            }
        }
        return -1;
    }
}

//[left,right]
class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        if (target < nums[0] || target > nums[right]) {
            return -1;
        }
        while(left <= right) {
            int mid = left + ((right-left)/2);
            if(nums[mid] == target){
                return mid;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else if (nums[mid] < target) {
                left = mid + 1;
            }
        }
        return -1;
    }
}

Double Pointers

Slow and Fast Pointer:

  • Fast Pointer: to look for the new Array’s element, new array is the array that does not contain target value
  • Slow Pointer: to index the updating index

27. Remove Element

Link

class Solution {
    public int removeElement(int[] nums, int val) {
        int size = nums.length;
        for(int i = 0; i < size; i++){
            if (nums[i] == val){ 
								// found the target value, move remaining element foward
                for (int j = i+1; j < size; j++){
                    nums[j-1] = nums[j]; 
                }
                i--; //since moved forward, we need to check first element
                size--;
            }
        }
        return size;
    }
}

nums = [0,1,2,2,3,0,4,2], val = 2

  • When fastIndex goes to nums[4] which nums[fastIndex] == val, slowIndex is still at nums[1]
  • Therefore, nums[2] == nums[4]
class Solution {
    public int removeElement(int[] nums, int val) {
        int slowIndex = 0;
        for (int fastIndex = 0; fastIndex < nums.length; fastIndex++) {
            if(nums[fastIndex] != val) {
                nums[slowIndex] = nums[fastIndex];
                slowIndex++;
            }
        }
        return slowIndex;
    }
}

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