Find if characters in string are present in another string
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I need to match if the characters in a string are present in a haystack with following conditions:
- If match string has repeated characters, haystack must contain multiple characters. In case of
hellohaystack must contain at-least 2l. - Character insensitive.
- Order insensitive.
The example below worked well for me as passed all the use cases. Can it be done in a better way with least complexity? Or maybe without built-in functions like indexOf, includes, substr. Though I am fine with map, filter like functions.
function array_diff(array1, array2) {
return array1.filter(e => array2.indexOf(e) === -1)
}
function compare(haystack, match) {
const haystackArr = haystack.toLowerCase().split('');
const matchArr = match.toLowerCase().split('');
return haystackArr.length >= array_diff(haystackArr, matchArr).length + matchArr.length;
}
console.log(compare('aldhedalfkjlhdlkjohd', 'hello'));javascript performance memory-optimization
asked 11 mins ago
Vikash Tiwari
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
I need to match if the characters in a string are present in a haystack with following conditions:
- If match string has repeated characters, haystack must contain multiple characters. In case of
hellohaystack must contain at-least 2l. - Character insensitive.
- Order insensitive.
The example below worked well for me as passed all the use cases. Can it be done in a better way with least complexity? Or maybe without built-in functions like indexOf, includes, substr. Though I am fine with map, filter like functions.
function array_diff(array1, array2) {
return array1.filter(e => array2.indexOf(e) === -1)
}
function compare(haystack, match) {
const haystackArr = haystack.toLowerCase().split('');
const matchArr = match.toLowerCase().split('');
return haystackArr.length >= array_diff(haystackArr, matchArr).length + matchArr.length;
}
console.log(compare('aldhedalfkjlhdlkjohd', 'hello'));javascript performance memory-optimization
asked 11 mins ago
Vikash Tiwari
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to match if the characters in a string are present in a haystack with following conditions:
- If match string has repeated characters, haystack must contain multiple characters. In case of
hellohaystack must contain at-least 2l. - Character insensitive.
- Order insensitive.
The example below worked well for me as passed all the use cases. Can it be done in a better way with least complexity? Or maybe without built-in functions like indexOf, includes, substr. Though I am fine with map, filter like functions.
function array_diff(array1, array2) {
return array1.filter(e => array2.indexOf(e) === -1)
}
function compare(haystack, match) {
const haystackArr = haystack.toLowerCase().split('');
const matchArr = match.toLowerCase().split('');
return haystackArr.length >= array_diff(haystackArr, matchArr).length + matchArr.length;
}
console.log(compare('aldhedalfkjlhdlkjohd', 'hello'));javascript performance memory-optimization
asked 11 mins ago
Vikash Tiwari
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need to match if the characters in a string are present in a haystack with following conditions:
- If match string has repeated characters, haystack must contain multiple characters. In case of
hellohaystack must contain at-least 2l. - Character insensitive.
- Order insensitive.
The example below worked well for me as passed all the use cases. Can it be done in a better way with least complexity? Or maybe without built-in functions like indexOf, includes, substr. Though I am fine with map, filter like functions.
function array_diff(array1, array2) {
return array1.filter(e => array2.indexOf(e) === -1)
}
function compare(haystack, match) {
const haystackArr = haystack.toLowerCase().split('');
const matchArr = match.toLowerCase().split('');
return haystackArr.length >= array_diff(haystackArr, matchArr).length + matchArr.length;
}
console.log(compare('aldhedalfkjlhdlkjohd', 'hello'));function array_diff(array1, array2) {
return array1.filter(e => array2.indexOf(e) === -1)
}
function compare(haystack, match) {
const haystackArr = haystack.toLowerCase().split('');
const matchArr = match.toLowerCase().split('');
return haystackArr.length >= array_diff(haystackArr, matchArr).length + matchArr.length;
}
console.log(compare('aldhedalfkjlhdlkjohd', 'hello'));function array_diff(array1, array2) {
return array1.filter(e => array2.indexOf(e) === -1)
}
function compare(haystack, match) {
const haystackArr = haystack.toLowerCase().split('');
const matchArr = match.toLowerCase().split('');
return haystackArr.length >= array_diff(haystackArr, matchArr).length + matchArr.length;
}
console.log(compare('aldhedalfkjlhdlkjohd', 'hello'));javascript performance memory-optimization
javascript performance memory-optimization
asked 11 mins ago
Vikash Tiwari
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
Vikash Tiwari
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
Vikash Tiwari
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 mins ago
Vikash Tiwari
1
asked 11 mins ago
Vikash Tiwari
1
1
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vikash Tiwari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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Vikash Tiwari is a new contributor. Be nice, and check out our Code of Conduct.
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