About Algorithm: Difference between revisions
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=Sorting Algoritms= | =Sorting Algoritms= | ||
* Two Pass - | * Two Pass - | ||
<pre class="code java"> | |||
class Solution { | class Solution { | ||
public int[] productExceptSelf(int[] nums) { | public int[] productExceptSelf(int[] nums) { | ||
| Line 30: | Line 31: | ||
} | } | ||
} | } | ||
</pre> | |||
* [[Index.php?title=Binary Search|Binary Search]] - Binary search exploit "sorted and rotated" structure. At every iteration, it divide the space by 2 using left or right. | * [[Index.php?title=Binary Search|Binary Search]] - Binary search exploit "sorted and rotated" structure. At every iteration, it divide the space by 2 using left or right. | ||
<pre class="code java"> | <pre class="code java"> | ||
Revision as of 23:00, 5 October 2025
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/
- Two pass are necessary to calculate the left and right products https://leetcode.com/problems/product-of-array-except-self/
- A product invert the sign, the max becomes the min, and the min becomes the max: https://leetcode.com/problems/maximum-product-subarray/
- Binary search exploit "sorted and rotated" structure. At every iteration, it divide the space by 2 using left or right. https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
- Search in Rotated Sorted Array https://leetcode.com/problems/search-in-rotated-sorted-array/
- Find the pivot
- Binary Search to left side, Binary Search to right side
Sorting Algoritms
- Two Pass -
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 - Binary search exploit "sorted and rotated" structure. At every iteration, it divide the space by 2 using left or right.
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;
}
- Fibonacci
- 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