About Algorithm

Tips

• Use Set/Hashet for unique values https://leetcode.com/problems/contains-duplicate/
• Use Math.max(int,int) to get the max number between a number and a calculation. https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
• A product invert the sign, the max becomes the min, and the min becomes the max: https://leetcode.com/problems/maximum-product-subarray/

Array Algoritms

Kadane’s Algorithm - adapted on cumulated products

    public int[] productExceptSelf(int[] nums) {
        int length = nums.length;
        int[] answer = new int[nums.length];

        int leftProduct = 1;
        for (int i = 0; i < length; i++) {
            answer[i] = leftProduct;
            leftProduct *= nums[i];
        }

        int rightProduct = 1;
        for (int i = length - 1; i >= 0; i--) {
            answer[i] *= rightProduct;
            rightProduct *= nums[i];
        }

        return answer;
    }

Two Pass - (LeetCode)- to calculate the left and right products

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int length = nums.length;
        int[] answer = new int[nums.length];

        int leftProduct = 1;
        for (int i = 0; i < length; i++) {
            answer[i] = leftProduct;
            leftProduct *= nums[i];
        }

        int rightProduct = 1;
        for (int i = length - 1; i >= 0; i--) {
            answer[i] *= rightProduct;
            rightProduct *= nums[i];
        }

        return answer;
    }
}

Binary Search - (LeetCode - LeetCode) - At every iteration, it divide the space by 2 using left or right.

• Find the pivot
• Binary Search to left side, Binary Search to right side

    private int binarySearch(
        int[] nums,
        int leftBoundary,
        int rightBoundary,
        int target
    ) {
        int leftIndex = leftBoundary;
        int rightIndex = rightBoundary;

        while (leftIndex <= rightIndex) {
            int midIndex = (leftIndex + rightIndex) / 2;
            int midValue = nums[midIndex];

            if (midValue == target) {
                return midIndex;
            } else if (midValue > target) {
                rightIndex = midIndex - 1;
            } else {
                leftIndex = midIndex + 1;
            }
        }
        return -1;
    }

MergeSort

• QuickSort (Java)
• Bit Manipulation
• Backtracking
• Graph Traversal
• Union-Find
• Dynamic Programming
• Greedy
• Sliding Window / Two Pointers
• BubbleSort (Java)
  • Swapping the adjacent numbers two by two by order

Graphs

• Dijkstra
• Topologic sorting

Cryptography & Compression

• Shannon-Fano
• Huffman
• Diffie-Hellman
• RSA

Prime Numbers

• ?

Most beautiful Equation

• P=NP (Les Équations de Yang Mills)
• Millenium problems (Clay Mathematics Institute)
• Few ideas