Merge branch 'master' into cycle-detection-in-linkedlist

This commit is contained in:
Omkarnath Parida
2020-10-04 23:50:56 +05:30
committed by GitHub
25 changed files with 8555 additions and 34 deletions

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/*
* You are climbing a stair case. It takes n steps to reach to the top.
* Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
*/
const climbStairs = (n) => {
let prev = 0
let cur = 1
let temp
for (let i = 0; i < n; i++) {
temp = prev
prev = cur
cur += temp
}
return cur
}
const main = () => {
const number = 5
console.log('Number of ways to climb ' + number + ' stairs in ' + climbStairs(5))
}
// testing
main()

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/*
Wikipedia -> https://en.wikipedia.org/wiki/Edit_distance
Q. -> Given two strings `word1` and `word2`. You can perform these operations on any of the string to make both strings similar.
- Insert
- Remove
- Replace
Find the minimum operation cost required to make both same. Each operation cost is 1.
Algorithm details ->
time complexity - O(n*m)
space complexity - O(n*m)
*/
const minimumEditDistance = (word1, word2) => {
const n = word1.length
const m = word2.length
const dp = new Array(m + 1).fill(0).map(item => [])
/*
fill dp matrix with default values -
- first row is filled considering no elements in word2.
- first column filled considering no elements in word1.
*/
for (let i = 0; i < n + 1; i++) {
dp[0][i] = i
}
for (let i = 0; i < m + 1; i++) {
dp[i][0] = i
}
/*
indexing is 1 based for dp matrix as we defined some known values at first row and first column/
*/
for (let i = 1; i < m + 1; i++) {
for (let j = 1; j < n + 1; j++) {
const letter1 = word1[j - 1]
const letter2 = word2[i - 1]
if (letter1 === letter2) {
dp[i][j] = dp[i - 1][j - 1]
} else {
dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j - 1], dp[i][j - 1]) + 1
}
}
}
return dp[m][n]
}
const main = () => {
console.log(minimumEditDistance('horse', 'ros'))
console.log(minimumEditDistance('cat', 'cut'))
console.log(minimumEditDistance('', 'abc'))
console.log(minimumEditDistance('google', 'glgool'))
}
main()