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25.DynamicProgramming.js
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25.DynamicProgramming.js
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// WHAT IS DYNAMIC PROGRAMMING
// "A method for solving a complex problem by breaking it down into a collection
// of simpler subproblems, solving each of those subproblems just once, and storing their solutions."
// "Using past knowledge to make solving a future problem easier"
// IT ONLY WORKS ON PROBLEMS WITH...
// OPTIMAL SUBSTRUCTURE &
// OVERLAPPING SUBPROBLEMS
// OVERLAPPING SUBPROBLEMS
// A problem is said to have overlapping subproblems if it can be broken down into subproblems which are reused several times
// e.g, Fibonacci Sequence
// OPTIMAL SUBSTRUCTURE
// A problem is said to have optimal substructure if an optimal solution can be constructed from optimal solutions of its subproblems
// recursive - O( 2^n ), more precisely O( 1.6^n ), exponential time complexity
// space complexity is O(n)
function fib(n){
if(n <= 2) return 1;
return fib(n-1) + fib(n-2);
}
// WHAT IS THE SUBSTRUCTURE?
// fib(n) = fib(n - 1) + fib(n - 2);
// Two Flavours of Dynamic Programming
// 1. MEMOIZATION - TOP-DOWN APPROACH
// Storing the results of expensive function calls and returning the cached result when the same inputs occur again
// memoized - O(n), linear time complexity
function fib_memo(n, memo=[]){
if(memo[n] !== undefined) return memo[n]
if(n <= 2) return 1;
let res = fib(n-1, memo) + fib(n-2, memo);
memo[n] = res;
return res;
}
// function fib_memo(n, memo=[undefined, 1, 1]){
// if(memo[n] !== undefined) return memo[n];
// var res = fib(n-1, memo) + fib(n-2, memo);
// memo[n] = res;
// return res;
// }
// function fib_memo(n, savedFib={}) {
// // base case
// if (n <= 0) { return 0; }
// if (n <= 2) { return 1; }
//
// // memoize
// if (savedFib[n - 1] === undefined) {
// savedFib[n - 1] = fib(n - 1, savedFib);
// }
//
// // memoize
// if (savedFib[n - 2] === undefined) {
// savedFib[n - 2] = fib(n - 2, savedFib);
// }
//
// return savedFib[n - 1] + savedFib[n - 2];
// }
// Say you are a traveller on a 2D grid. You start at the top left corner and your goal
// is to travel to the bottom right corner. You can only move down or right.
// In how many ways can can you travel to the goal on a grid of width m * n;
// O(2^ n+m) time, O(n+m) space
function gridTraveler(m, n) {
if(m === 1 && n === 1) return 1;
if(m === 0 || n === 0) return 0;
return gridTraveler(m-1, n) + gridTraveler(m, n-1)
// m-1 represents going down, n-1 represents going right
}
function gridTravelerMemo(m, n, memo={}) {
let key = `${m},${n}`;
if(key in memo) return memo[key];
if(m === 1 && n === 1) return 1;
if(m === 0 || n === 0) return 0;
memo[key] = gridTraveler(m-1, n, memo) + gridTraveler(m, n-1, memo);
return memo[key]
}
gridTraveler(3, 3) // 6
function canConstruct(target, wordBank, memo={}) {
if(target in memo) return memo[target];
if(target === "") return true;
for (let word of wordBank) {
if(target.indexOf(word) === 0) {
let suffix = target.slice(word.length);
if(canConstruct(suffix, wordBank, memo) === true) {
memo[targrt] = true;
return true
}
}
}
memo[target] = false
return memo[target];
}
canConstruct("abcdef", ["ab", "abc", "cd", "def", "abcd"]) // true;
// 2. TABULATION - BOTTOM-UP APPROACH
// Storing the result of a previous result in a "table" (usually an array)
// Usually done using iteration
// Better space complexity can be achieved using tabulation
// tabulated
function fib_table(n){
if(n <= 2) return 1;
var fibNums = [0,1,1];
for(var i = 3; i <= n; i++){
fibNums[i] = fibNums[i-1] + fibNums[i-2];
}
return fibNums[n];
}
// function fib_table(n){
// const table = Array(n+1).fill(0)
// table[1] = 1
// for(var i = 0; i <= n; i++){
// table[i + 1] += table[i];
// table[i + 2] += table[i];
// }
// return table[n];
// }
// function fib_table(n){
// const lastTwo = [0, 1];
// let counter = 3;
//
// while (counter <= n) {
// const nextFib = last[0] + last[1];
//
// lastTwo[0] = lastTwo[1];
// lastTwo[1] = nextFib;
//
// counter++
// }
//
// return n > 1 ? lastTwo[1] : lastTwo[0];
// }
// let calculations = 0;
// function fibonacci(n) { //O(2^n)
// if (n < 2) {
// return n
// }
// return fibonacci(n-1) + fibonacci(n-2);
// }
// function fibonacciMaster() { //O(n)
// let cache = {};
// return function fib(n) {
// calculations++;
// if (n in cache) {
// return cache[n];
// } else {
// if (n < 2) {
// return n;
// } else {
// cache[n] = fib(n-1) + fib(n-2);
// return cache[n];
// }
// }
// }
// }
// function fibonacciMaster2(n) {
// let answer = [0,1];
// for ( let i = 2; i <= n; i++) {
// answer.push(answer[i-2]+ answer[i-1]);
// }
// return answer.pop();
// }
// const fasterFib = fibonacciMaster();
// console.log('Slow', fibonacci(35))
// console.log('DP', fasterFib(100));
// console.log('DP2', fibonacciMaster2(100));
// console.log('we did ' + calculations + ' calculations');
// LCS (Longest Common subsequence)
// MEMOIZATION, TOP DOWN APPROACH
let longestCommonSubsequence = function(text1, text2) {
function dp(i, j, memo={}) {
if(`${i},${j}` in memo) {
return memo[`${i},${j}`]
}
if(i === -1 || j === -1) {
return 0;
}
if(text1[i] === text2[j]) {
memo[`${i},${j}`] = 1 + dp(i-1, j-1, memo);
return memo[`${i},${j}`]
} else {
memo[`${i},${j}`] = Math.max(dp(i-1, j, memo), dp(i, j-1, memo))
return memo[`${i},${j}`]
}
}
return dp(text1.length-1, text2.length-1)
};
// TABULATION, BOTTOM UP APPROACH
let longestCommonSubsequenceTabulated = function(text1, text2) {
text1 = " " + text1;
text2 = " " + text2;
let width = text1.length;
let height = text2.length;
let table = [...Array(height)].map(x => [...Array(width)].map(ele => 0))
for(let y = 1; y < height; y++) {
for(let x = 1; x < width; x++) {
if(text1[x] === text2[y]) {
// with the same character
// extend the length of common subsequence
table[y][x] = 1 + table[y-1][x-1]
} else {
// with different characters
// choose the optimal subsequence
table[y][x] = Math.max(table[y-1][x], table[y][x-1])
}
}
}
return table[height-1][width-1]
};
longestCommonSubsequence("abcde", "ace") // 3