Add Strings

class Solution {
    public String addStrings(String num1, String num2) {
        int size1 = num1.length();
        int size2 = num2.length();
        int p1 = size1-1;
        int p2 = size2-1;
        StringBuilder out = new StringBuilder();
        int c = 0;
        while (p1>=0 || p2>=0) {
            int n1 = 0;
            int n2 = 0;
            if (p1>=0) {
                n1 = num1.charAt(p1--)-'0';
            }
            if (p2>=0) {
                n2 = num2.charAt(p2--)-'0';
            }
            int temp = n1+n2+c;
            out.insert(0,String.valueOf(temp%10));
            c = temp>=10? 1:0;
        }
        if (c>0) out.insert(0,String.valueOf(1)); // add the final carry out
        return out.toString();
    }
}

Reverse Integer

LeetCode Problem #7

Java Solution:

class Solution {
    public int reverse(int x) {
        boolean neg = false;
        if (x<0) {
            x= -x;
            neg = true;
        }
        long out = 0;
        while (x>0) {
            int last = x%10;
            x = x/10;
            out = out * 10 + last;
        }
        if (out>Integer.MAX_VALUE) return 0;
        if (neg) return (int)-out;
        return (int)out;
    }
}

Python Solution:

class Solution:
    def reverse(self, x: int) -> int:
        inX = abs(x)
        out = 0
        while (inX>0):
            out = out * 10 + inX % 10
            inX = inX // 10
        if out not in range(-2**31,2**31):
            return 0
        if (x > 0):
            return out
        else:
            return -out

2022-10-06

Maximum Product of Three Numbers

LeetCode Problem #628

class Solution {
    public int maximumProduct(int[] nums) {
        
        // M1 Sorting
        int size = nums.length;
        Arrays.sort(nums);
        return Math.max(nums[0]*nums[1]*nums[size-1], nums[size-3]*nums[size-2]*nums[size-1]);
        
    }
}

class Solution {
    public int maximumProduct(int[] nums) {
        
        // M2 Looping
        int min1 = Integer.MAX_VALUE; // the smallest
        int min2 = Integer.MAX_VALUE; // the second smallest
        
        int max1 = Integer.MIN_VALUE; // the largest
        int max2 = Integer.MIN_VALUE; // the second largest
        int max3 = Integer.MIN_VALUE; // the third largest
        
        for (int n:nums) {
            if (n<=min1) { // n become the new smallest
                min2 = min1;
                min1 = n;
            } else if (n<=min2) { // n become the new second smallest
                min2 = n;
            }
            
            if (n>=max1) { // n become the new largest
                max3 = max2;
                max2 = max1;
                max1 = n;
            }
            else if (n>=max2) { // n become the new second largest
                max3 = max2;
                max2 = n;
            }
            else if (n>=max3) { // n become the new third largest
                max3 = n;
            }
        }
        
        return Math.max(min1*min2*max1, max3*max2*max1);    
    }
}