问题
The challenge is next:
At a job interview, you are challenged to write an algorithm to check if a given string, s, can be formed from two other strings, part1 and part2. The restriction is that the characters in part1 and part2 are in the same order as in s. The interviewer gives you the following example and tells you to figure out the rest from the given test cases.
What am I doing wrong? Why it fails anyway?
I wrote 2 different scripts, and both aren't working for some test cases
function isMerge(s, part1, part2) {
var sArr = s.split('');
var part1Arr = part1.split('');
var part2Arr = part2.split('');
var tempArr = new Array(sArr.length);
function compareArrays(arr1, arr2){
var count = 0;
for (var i = 0; i < arr1.length; i++) {
if (arr1[i] !== arr2[i]) count++;
}
return (count == 0);
}
for (var i = 0; i < sArr.length; i++) {
for (var j = 0; j < part1Arr.length; j++) {
if (sArr[i] == part1Arr[j]) tempArr[i] = j;
}
for (var k = 0; k < part2Arr.length; k++) {
if (sArr[i] == part2Arr[k]) tempArr[i] = k;
}
}
alert(tempArr);
var check = tempArr.slice();
check.sort();
alert(check);
if (compareArrays(tempArr, check)) return true;
else return false;
}
alert(isMerge('codewars', 'cdw', 'oears'));function isMerge(s, part1, part2) {
// create arrays of letters
var sArr = s.split('');
var part1Arr = part1.split('');
var part2Arr = part2.split('');
// create an associative array 'temp' (0:C, 1:O and so on)
var temp = {};
for (var k = 0; k < sArr.length; k++) {
temp[k] = sArr[k];
}
// reverse an associative array 'temp' (now C:0, O:0 and so on)
for (var key in temp) {
var keyTemp = key;
var keyValue = temp[key];
key = keyValue;
temp[key] = keyTemp;
}
// the function that compares arrays
function compareArrays(arr1, arr2){
var count = 0;
for (var i = 0; i < arr1.length; i++) {
if (arr1[i] !== arr2[i]) count++;
}
return (count == 0);
}
// sorting function
function order(a, b) {
var comparingA;
var comparingB;
for (var char in temp) {
if (char == a) {
comparingA = temp[char]; // comparingA is the number of 'a' in object 'temp'
}
if (char == b){
comparingB = temp[char]; // comparingB is the number of 'b' in object 'temp'
}
}
return (comparingA - comparingB);
}
// create copies of arrays
var part1Sorted = part1Arr.slice();
var part2Sorted = part2Arr.slice();
// and sort that copies
part1Sorted.sort(order);
part2Sorted.sort(order);
// If the array did not change after sorting, the order of the letters was correct
if (compareArrays(part1Sorted, part1Arr) && compareArrays(part2Sorted, part2Arr)) {
// so now we can check is merge possible
sArr = sArr.sort();
var parts = part1Arr.concat(part2Arr);
parts = parts.sort();
var res = compareArrays(sArr, parts);
return res;
}
return false;
}
alert(isMerge('codewars', 'code', 'wasr'));
alert(isMerge('codewars', 'oers', 'cdwa'));I just added comments to the second script
回答1:
I find it hard to understand what your code is attempting to do. It would help if you provided comments as well as explain the idea behind the algorithm/s you are attempting to implement.
Here's a commented example of a recursion that considers if pointers i and j to parts 1 & 2 could constitute a valid merge up to that point.
function isMerge(s, part1, part2) {
// Merge is invalid if the parts' lengths don't add up to the string's
if (part1.length + part2.length != s.length)
return false;
// Recursive function
function g(i, j){
// Base case: both pointers are exactly at the end of each part
if (i == part1.length && j == part2.length)
return true;
// One of our pointers has extended beyond the part's length,
// that couldn't be right
if (i > part1.length || j > part2.length)
return false;
// Just part1 matches here so increment i
if (part1[i] == s[i + j] && part2[j] != s[i + j])
return g(i + 1, j);
// Just part2 matches here so increment j
else if (part1[i] != s[i + j] && part2[j] == s[i + j])
return g(i, j + 1);
// Both parts match here so try incrementing either pointer
// to see if one of those solutions is correct
else if (part1[i] == s[i + j] && part2[j] == s[i + j])
return g(i + 1, j) || g(i, j + 1);
// Neither part matches here
return false;
}
// Call the recursive function
return g(0,0);
}
console.log(isMerge('codewars', 'cdw', 'oears'));
console.log(isMerge('codecoda', 'coda', 'code'));
console.log(isMerge('codewars', 'oers', 'cdwa'));
console.log(isMerge('codewars', 'cdw', 'years'));Stack version for really long strings:
function isMerge2(s, part1, part2) {
if (part1.length + part2.length != s.length)
return false;
let stack = [[0,0]];
while (stack.length){
[i, j] = stack.pop();
if (i == part1.length && j == part2.length)
return true;
if (i > part1.length || j > part2.length)
continue;
if (part1[i] == s[i + j] && part2[j] != s[i + j])
stack.push([i + 1, j]);
else if (part1[i] != s[i + j] && part2[j] == s[i + j])
stack.push([i, j + 1]);
else if (part1[i] == s[i + j] && part2[j] == s[i + j]){
stack.push([i + 1, j]);
stack.push([i, j + 1]);
}
}
return false;
}
function test(){
let s = '';
for (let i=0; i<1000000; i++)
s += ['a','b','c','d','e','f','g'][~~(Math.random()*6)];
let lr = {
l: '',
r: ''
};
for (let i=0; i<s.length; i++){
let which = ['l', 'r'][~~(Math.random()*2)];
lr[which] += s[i];
}
console.log(isMerge2(s,lr.l,lr.r));
}
test();回答2:
This is a recursive approach: it checks for a match of the first character of the string with either of the parts, and if there's a match recurses to try and match the rest of the string with the rest of the parts. The tricky thing is that when the first character of both parts is the same, you have to check if you can match against either of them (this solves the Bananas test).
function isMerge(str, p1, p2) {
if (!str.length) return !p1.length && !p2.length;
if (p1.length && str.charAt(0) == p1.charAt(0)) {
if (p2.length && str.charAt(0) == p2.charAt(0)) {
return isMerge(str.substr(1), p1.substr(1), p2) || isMerge(str.substr(1), p1, p2.substr(1));
}
else {
return isMerge(str.substr(1), p1.substr(1), p2);
}
}
else if (p2.length && str.charAt(0) == p2.charAt(0)) {
return isMerge(str.substr(1), p1, p2.substr(1));
}
else {
return false;
}
}
回答3:
Recursive Call:
You could use a recursive approach by first checking the length of string and part strings and then by the length or by checking a character and by checking the rest of the given part strings.
function isMerge(s, part1, part2) {
if (s.length !== part1.length + part2.length) {
return false;
}
if (!s.length) {
return true;
}
if (part1[0] === s[0] && isMerge(s.slice(1), part1.slice(1), part2)) {
return true;
}
if (part2[0] === s[0] && isMerge(s.slice(1), part1, part2.slice(1))) {
return true;
}
return false;
}
console.log(isMerge('codewars', 'cdw', 'oears')); // true
console.log(isMerge('codewars', 'oers', 'cdwa')); // true
console.log(isMerge('codecoda', 'coda', 'code')); // true
console.log(isMerge('baeabb', 'b', 'baeab')); // true
console.log(isMerge('bdab', 'bdab', '')); // true
console.log(isMerge('bfaef', 'f', 'bfae')); // true
console.log(isMerge('codewars', 'cdw', 'years')); // false
console.log(isMerge('codewars', 'code', 'warss')); // false
console.log(isMerge('codewars', 'codes', 'wars')); // false
console.log(isMerge('', 'a', 'b')); // false.as-console-wrapper { max-height: 100% !important; top: 0; }With a Stack instead of a Recursive Call:
function isMerge(s, part1, part2) {
var stack = [[s, part1, part2]];
if (s.length !== part1.length + part2.length) {
return false;
}
while (stack.length) {
[s, part1, part2] = stack.shift();
if (!s.length) {
return true;
}
if (part1[0] === s[0]) {
stack.push([s.slice(1), part1.slice(1), part2]);
}
if (part2[0] === s[0]) {
stack.push([s.slice(1), part1, part2.slice(1)]);
}
}
return false;
}
console.log(isMerge('codewars', 'cdw', 'oears')); // true
console.log(isMerge('codewars', 'oers', 'cdwa')); // true
console.log(isMerge('codecoda', 'coda', 'code')); // true
console.log(isMerge('baeabb', 'b', 'baeab')); // true
console.log(isMerge('bdab', 'bdab', '')); // true
console.log(isMerge('bfaef', 'f', 'bfae')); // true
console.log(isMerge('codewars', 'cdw', 'years')); // false
console.log(isMerge('codewars', 'code', 'warss')); // false
console.log(isMerge('codewars', 'codes', 'wars')); // false
console.log(isMerge('', 'a', 'b')); // false.as-console-wrapper { max-height: 100% !important; top: 0; }回答4:
May you try this below?
function isMerge(s, part1, part2) {
var result= true;
var total = part1 + part2;
for (var i = 0; i < s.length; i++) {
var char = s.charAt(i);
if(total.indexOf(char) === -1) {
result = false;
break;
}
}
return result;
}
回答5:
First of all, here is your code just working:
function isMerge(s, part1, part2) {
var sArr = s.split('');
var part1Arr = part1.split('');
var part2Arr = part2.split('');
var tempArr = new Array(sArr.length);
function compareArrays(arr1, arr2){
var count = 0;
for (var i = 0; i < arr1.length; i++) {
if (arr1[i] != arr2[i]) count++;
}
return (count == 0);
}
for (var i = 0; i < sArr.length; i++) {
for (var j = 0; j < part1Arr.length; j++) {
if (sArr[i] == part1Arr[j]) tempArr[i] = part1Arr[j];
}
for (var k = 0; k < part2Arr.length; k++) {
if (sArr[i] == part2Arr[k]) tempArr[i] = part2Arr[k];
}
}
alert(tempArr);
if (compareArrays(tempArr, sArr)) return true;
else return false;
}
alert(isMerge('codewars', 'cdw', 'oears'));
Now, what was the problem?
for (var j = 0; j < part1Arr.length; j++) {
/* Here you assigned the index (tempArr[i] = j;) not the char */
if (sArr[i] == part1Arr[j]) tempArr[i] = part1Arr[j];
}
for (var k = 0; k < part2Arr.length; k++) {
/* Here you assigned the index (tempArr[i] = k;) not the char */
if (sArr[i] == part2Arr[k]) tempArr[i] = part2Arr[k];
}
Hope i could help you ;)
回答6:
Simple merge logic:
function isMerge(s, part1, part2) {
var sArr = s.split('');
var part1Arr = part1.split('');
var part2Arr = part2.split('');
var j = 0;
var k = 0;
for (var i = 0; i < sArr.length; i++) {
if ((j < part1Arr.length) && (sArr[i] == part1Arr[j]))
j++
else if ((k < part2Arr.length) && (sArr[i] == part2Arr[k]))
k++
else
break
}
return (j == part1Arr.length && k == part2Arr.length && (j + k) == sArr.length);
}
console.log(isMerge('abcd', 'ac', 'bd'));
console.log(isMerge('abcd', 'acd', 'b'));
console.log(isMerge('abcd', 'ac', 'b'));
console.log(isMerge('abcd', 'ac', 'db'));
console.log(isMerge('abcd', 'c', 'b'));
console.log(isMerge('a', '', 'a'));
console.log(isMerge('', '', ''));
console.log(isMerge('', 'a', 'b'));
console.log(isMerge('ab', '', ''));来源：https://stackoverflow.com/questions/49960458/codewars-merged-string-checker