If the first element of the array contains all the letters of the second element, then return true

Question

If the first element of the array contains all the letters of the second element, then return true

I attempted to tackle this challenge by creating a loop that iterates through the firstword each time a letter matches with a letter in the secondword.

function mutation(arr) { 
  var compare = [];
  var firstword = arr[0].toLowerCase();
  var secondword = arr[1].toLowerCase();
  var j = 0;
  for (let i = 0; i < firstword.length; i++) {

      if (firstword[i] === secondword[j]) {
      compare.push(secondword[i]);
      i = -1;
      j++;

      
      
      
    }
  }
  let result = compare.join("")
  if (result.length === secondword.length) {
    return true;
    
  } else {
    return false;
  }
}

console.log(mutation(["Noel", "Ole"]));

While this approach works in some scenarios, it fails in cases like the one above. What could be causing this inconsistency?

javascript arrays loops

Answer 1

Answer №1

Ensure you compare.push(secondword[j]) rather than compare.push(secondword[i])

function checkMutation(arr) {
  var compare = [];
  var firstword = arr[0].toLowerCase();
  var secondword = arr[1].toLowerCase();
  var j = 0;
  for (let i = 0; i < firstword.length; i++) {
    if (firstword[i] === secondword[j]) {
      compare.push(secondword[j]); // Correction here
      i = -1;
      j++;
    }
  }
  let result = compare.join("");
  if (result.length === secondword.length) {
    return true;
  } else {
    return false;
  }
}

console.log(checkMutation(["Noel", "Ole"]));

Additionally, you may want to explore Array.prototype.every.

const checkMutation = ([first, sec]) => {
  const lowerCaseFirst = first.toLowerCase();
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => lowerCaseFirst.includes(ch));
};

console.log(checkMutation(["Noel", "Ole"]));

If the strings are small, String.prototype.includes is suitable, but for larger strings, consider utilizing a Set.

const checkMutation = ([first, sec]) => {
  const firstSet = new Set(first.toLowerCase());
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => firstSet.has(ch));
};

console.log(checkMutation(["Noel", "Ole"]));

Answer 2

Ensure you compare.push(secondword[j]) rather than compare.push(secondword[i])

function checkMutation(arr) {
  var compare = [];
  var firstword = arr[0].toLowerCase();
  var secondword = arr[1].toLowerCase();
  var j = 0;
  for (let i = 0; i < firstword.length; i++) {
    if (firstword[i] === secondword[j]) {
      compare.push(secondword[j]); // Correction here
      i = -1;
      j++;
    }
  }
  let result = compare.join("");
  if (result.length === secondword.length) {
    return true;
  } else {
    return false;
  }
}

console.log(checkMutation(["Noel", "Ole"]));

Additionally, you may want to explore Array.prototype.every.

const checkMutation = ([first, sec]) => {
  const lowerCaseFirst = first.toLowerCase();
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => lowerCaseFirst.includes(ch));
};

console.log(checkMutation(["Noel", "Ole"]));

If the strings are small, String.prototype.includes is suitable, but for larger strings, consider utilizing a Set.

const checkMutation = ([first, sec]) => {
  const firstSet = new Set(first.toLowerCase());
  const lowerCaseSec = sec.toLowerCase();
  return Array.from(lowerCaseSec).every((ch) => firstSet.has(ch));
};

console.log(checkMutation(["Noel", "Ole"]));

Answer 3

Answer №2

Here is a concise ES6 function that utilizes the .every() method to check if every character in the secondword is present in the firstword. It will return true if all characters are found.

function characterCheck(arr) {
          const firstword = arr[0].toLowerCase();
          const secondword = arr[1].toLowerCase();
          return secondword.split('').every(char => firstword.includes(char));
}

console.log(characterCheck(["Noel", "Ole"]));

Answer 4

Here is a concise ES6 function that utilizes the .every() method to check if every character in the secondword is present in the firstword. It will return true if all characters are found.

function characterCheck(arr) {
          const firstword = arr[0].toLowerCase();
          const secondword = arr[1].toLowerCase();
          return secondword.split('').every(char => firstword.includes(char));
}

console.log(characterCheck(["Noel", "Ole"]));

Answer 5

Answer №3

Utilizing a Set in the provided solution is effective if there are no duplicate characters to consider in the second string. However, an alternative approach involves employing a Map to keep track of character counts. In this scenario, the map's key represents the character from the first string, while the corresponding value indicates the frequency of occurrence. This method ensures that scenarios like having

["Noel", "Ole"]

return true, whereas

["Noel", "Olle"]

would yield false, as the first string lacks sufficient "l" characters. The algorithm involves iterating through the second string and decrementing the character counts accordingly. If a character is missing or its count drops below 1 in the map, the function promptly returns false.

function mutation(arr: string[]): boolean {
    return s1ContainsAllCharsInS2(arr[0].toLowerCase(), arr[1].toLowerCase());
}

function s1ContainsAllCharsInS2(s1: string, s2: string): boolean {
    if (s2.length > s1.length) {
        return false;
    }
    let charCountMap: Map<string, number> = new Map<string, number>();
    Array.from(s1).forEach(c => {
        let currentCharCount: number = charCountMap.get(c);
        charCountMap.set(c, 1 + (currentCharCount ? currentCharCount : 0));
    });
    return !Array.from(s2).some(c => {
        let currentCharCount: number = charCountMap.get(c);
        if (!currentCharCount || currentCharCount < 1){
            return true;
        }
        charCountMap.set(c, currentCharCount - 1);
    });
}

Answer 6

Utilizing a Set in the provided solution is effective if there are no duplicate characters to consider in the second string. However, an alternative approach involves employing a Map to keep track of character counts. In this scenario, the map's key represents the character from the first string, while the corresponding value indicates the frequency of occurrence. This method ensures that scenarios like having

["Noel", "Ole"]

return true, whereas

["Noel", "Olle"]

would yield false, as the first string lacks sufficient "l" characters. The algorithm involves iterating through the second string and decrementing the character counts accordingly. If a character is missing or its count drops below 1 in the map, the function promptly returns false.

function mutation(arr: string[]): boolean {
    return s1ContainsAllCharsInS2(arr[0].toLowerCase(), arr[1].toLowerCase());
}

function s1ContainsAllCharsInS2(s1: string, s2: string): boolean {
    if (s2.length > s1.length) {
        return false;
    }
    let charCountMap: Map<string, number> = new Map<string, number>();
    Array.from(s1).forEach(c => {
        let currentCharCount: number = charCountMap.get(c);
        charCountMap.set(c, 1 + (currentCharCount ? currentCharCount : 0));
    });
    return !Array.from(s2).some(c => {
        let currentCharCount: number = charCountMap.get(c);
        if (!currentCharCount || currentCharCount < 1){
            return true;
        }
        charCountMap.set(c, currentCharCount - 1);
    });
}

Answer 7

Answer №4

An alternative method.

Creating a character map and comparing against it.

function compareCharacters(arr) {
  const charMap = {};

  for (let char of arr[0].toLowerCase()) {
    charMap[char] = true;
  }

  for (let char of arr[1].toLowerCase()) {
    if (!charMap[char]) {
      return false;
    }
  }

  return true;
}

console.log(compareCharacters(["Noel", "Ole"]));
console.log(compareCharacters(["Noel", "Oleeeeeeeeeeeee"]));

If the character count is important (your current code does not consider it), you can calculate the occurrences of each character and compare these counts.

function compareCharacters(arr) {
  const charMap = {};

  for (let char of arr[0].toLowerCase()) {
    charMap[char] = (charMap[char] || 0) + 1;
  }

  for (let char of arr[1].toLowerCase()) {
    if (!charMap[char]--) {
      return false;
    }
  }

  return true;
}

console.log(compareCharacters(["Noel", "Ole"]));
console.log(compareCharacters(["Noel", "Oleeeeeeeeeeeee"]));

Answer 8

An alternative method.

Creating a character map and comparing against it.

function compareCharacters(arr) {
  const charMap = {};

  for (let char of arr[0].toLowerCase()) {
    charMap[char] = true;
  }

  for (let char of arr[1].toLowerCase()) {
    if (!charMap[char]) {
      return false;
    }
  }

  return true;
}

console.log(compareCharacters(["Noel", "Ole"]));
console.log(compareCharacters(["Noel", "Oleeeeeeeeeeeee"]));

If the character count is important (your current code does not consider it), you can calculate the occurrences of each character and compare these counts.

function compareCharacters(arr) {
  const charMap = {};

  for (let char of arr[0].toLowerCase()) {
    charMap[char] = (charMap[char] || 0) + 1;
  }

  for (let char of arr[1].toLowerCase()) {
    if (!charMap[char]--) {
      return false;
    }
  }

  return true;
}

console.log(compareCharacters(["Noel", "Ole"]));
console.log(compareCharacters(["Noel", "Oleeeeeeeeeeeee"]));

If the first element of the array contains all the letters of the second element, then return true

Answer №1

Answer №2

Answer №3

Answer №4

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