Algorithm mutations - ashish9342/FreeCodeCamp GitHub Wiki
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- Return true if the string in the first element of the array contains all of the letters of the string in the second element of the array..
- If everything is lowercase it will be easier to compare.
try to solve the problem now
- Our strings might be easier to work with if they were arrays of characters.
try to solve the problem now
- A loop might help. Use
indexOf()
to check if the letter of the second word is on the first.
try to solve the problem now
Solution ahead!
Procedural
function mutation(arr) {
var test = arr[1].toLowerCase();
var target = arr[0].toLowerCase();
for (i=0;i<test.length;i++) {
if (target.indexOf(test[i]) < 0)
return false;
}
return true;
}
🚀 Run Code
First we make the two strings in the array lowercase. test
will hold what we are looking for in target
.
Then we loop through our test characters and if any of them is not found we return false
.
If they are all found, the loop will finish without returning anything and we get to return true
.
Declarative
function mutation(arr) {
return arr[1].toLowerCase()
.split('')
.every(function(letter) {
return arr[0].toLowerCase()
.indexOf(letter) != -1;
});
}
🚀 Run Code
Grab the second string, lowercase and turn it into an array; then make sure every one of its letters is a part of the lowercased first string.
Every
will basically give you letter by letter to compare, which we do by using indexOf
on the first string. indexOf
will give you -1 if the current letter
is missing. We check that not to be the case, for if this happens even once every
will be false.
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