CODEWITHSUNDEEP
javascript
RRR uzumaki
RRR uzumaki

Reputation: 1328

Removing last vowels from every word in a sentence javascript

Write a function that removes the last vowel in each word in a sentence.

Examples:

removeLastVowel("Those who dare to fail miserably can achieve greatly.") 
 ➞ "Thos wh dar t fal miserbly cn achiev gretly."

removeLastVowel("Love is a serious mental disease.") 
 ➞ "Lov s  serios mentl diseas"

removeLastVowel("Get busy living or get busy dying.") 
 ➞ "Gt bsy livng r gt bsy dyng"

What I am doing is

  function sumDigProd(arr) {
  let ans = arr.split(" ");
  let array = [];
  for (i = 0; i < ans.length; i++) {
    for (j = 0; j < ans[i].length; j++) {
      var vowelAtLast;
      if (
        ans[i][j].toLowerCase() == "a" ||
        ans[i][j].toLowerCase() == "e" ||
        ans[i][j].toLowerCase() == "i" ||
        ans[i][j].toLowerCase() == "o" ||
        ans[i][j].toLowerCase() == "u"
      ) {
        vowelAtLast = ans[i][j];
      }
    }
    var idex = ans[i].lastIndexOf(vowelAtLast);
    console.log(idex,ans[i],ans[i][idex]);
    ans[i].replace(ans[i][idex],'')
    array.push(ans[i])
    console.log(ans)
  }
  console.log(ans)
  return array.join(" ");
}

console.log(sumDigProd("youuo goalo people."));

Here, console.log(idex,ans[i],ans[i][idex]); gives me the correct output for example:

4 "youuo" "o"

But now when I try to do: 

ans[i].replace(ans[i][idex],'')
console.log(ans)

I get 

["youuo", "goalo", "people."]

again instead i should get

["youu", "goalo", "people."]

after the first time loop runs... and then at last I get the output as

["youu", "goal", "peopl."]

as per the problem but I get 

["youuo", "goalo", "people."]

why are no changes made by the replace method here?

Upvotes: 2

Views: 743

Answers (1)

CertainPerformance
CertainPerformance

Reputation: 371233

A problem is that

ans[i].replace(ans[i][idex], '')

will only replace the first occurrence of whatever character ans[i][idex] is. Eg

aza

would result in

za

Another issue is that you must use the return value of .replace, else it'll go unused and be irrelevant; you'd want something like

ans[i] = ans[i].replace(ans[i][idex], '')

instead, so that the item at that index in the array is properly reassigned.

But it would probably be easier to use a regular expression: match a vowel, followed by capturing zero or more non-vowels in a capture group, with lookahead matching a space or the end of the string. Then replace with the first capture group, thereby removing the last vowel:

const sumDigProd = str => str.replace(
  /[aeiou]([^aeiou]*?)(?= |$)/gi,
  '$1'
);

console.log(sumDigProd("youuo goalo people."));

[aeiou]([^aeiou]*?)(?= |$) means:

  • [aeiou] - Any vowel
  • ([^aeiou]*?) - Match and capture:
    • [^aeiou]*? Any non-vowel, lazily (capture group), up until
  • (?= |$) - lookahead matches a space or the end of the string

Then

  • '$1' - Replace with the capture group

To change your original code, identify the last index of a vowel by iterating from the final index of the string and going backwards. When one is found, reassign the string at ans[i] and .sliceing the portions behind and in front of the found vowel:

function sumDigProd(arr) {
  let ans = arr.split(" ");
  for (i = 0; i < ans.length; i++) {
    for (j = ans[i].length - 1; j >= 0; j--) {
      if (
        ans[i][j].toLowerCase() == "a" ||
        ans[i][j].toLowerCase() == "e" ||
        ans[i][j].toLowerCase() == "i" ||
        ans[i][j].toLowerCase() == "o" ||
        ans[i][j].toLowerCase() == "u"
      ) {
        ans[i] = ans[i].slice(0, j) + ans[i].slice(j + 1);
        // go to next word
        break;
      }
    }
  }
  return ans.join(' ');
}

console.log(sumDigProd("youuo goalo people."));

Upvotes: 4

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