Transform all characters in a string to uppercase sequentially and then save them in an array

Question

Transform all characters in a string to uppercase sequentially and then save them in an array

// Let's create a fun and unique function that transforms a simple string into an exciting Mexican Wave!
var smallarr=[]
var   bigarr=[]
var   another=""

function wave(str){
    for (var i=0;i<str.length;i++){
        smallarr.push(str)
    }
    for (var j=0;j<smallarr.length;j++)
    {
        if(smallarr[j][j].toUpperCase()==smallarr[j][j])
        { 
            var c=smallarr[j][j].toLowerCase()
            smallarr[j][j]=c
        }
        else {
            var c=smallarr[j][j].toUpperCase()
            smallarr[j][j]=c}
        }        
    }
    return smallarr     
}

document.write(wave("edabit"))
//console.log(smallarr)

Instead of a regular wave, the original output was not as expected. For example: wave("edabit") ➞ ["Edabit", "eDabit", "edAbit", "edaBit", "edabIt", "edabiT"] but it returned the same as the input.

If you have any suggestions or advice on improving this function, feel free to share!

javascript arrays multidimensional-array

Answer 1

Answer №1

To transform the input string, you can divide it into individual characters and then iterate over each element to replace it with its uppercase equivalent:

const convertToWave = (str) => {
  return str
    .split('')
    .map((char, index) =>
      str.substr(0, index) + char.toUpperCase() + str.substr(index + 1))
}

console.log(convertToWave('hello'));

Answer 2

To transform the input string, you can divide it into individual characters and then iterate over each element to replace it with its uppercase equivalent:

const convertToWave = (str) => {
  return str
    .split('')
    .map((char, index) =>
      str.substr(0, index) + char.toUpperCase() + str.substr(index + 1))
}

console.log(convertToWave('hello'));

Answer 3

Answer №2

If you want to transform a string into an array, you can utilize the Array.from() method. Within the callback function (mapFn), extract the current letter and index, then use them along with String.substring() to construct a new string with the capitalized letter.

const alterString = str => Array.from(str, (char, index) =>
  str.substring(0, index) + char.toUpperCase() + str.substring(index + 1)
);

const updatedStr = alterString('hello');

console.log(updatedStr);

Answer 4

If you want to transform a string into an array, you can utilize the Array.from() method. Within the callback function (mapFn), extract the current letter and index, then use them along with String.substring() to construct a new string with the capitalized letter.

const alterString = str => Array.from(str, (char, index) =>
  str.substring(0, index) + char.toUpperCase() + str.substring(index + 1)
);

const updatedStr = alterString('hello');

console.log(updatedStr);

Answer 5

Answer №3

If you want to experiment with Array.map(), consider using the second callback parameter to represent an index:

let input = "coding";

let result = input.split('').map(
     (_, i) => input.split('').map(
          (char, j) => i === j ? char.toUpperCase() : char).join()
 );

console.log(result);

UPDATE:

An issue with your method is that strings are immutable, so a new string should be constructed using slice as shown below:

var smallarr=[]
   var bigarr=[]
   var another=""
   function wave(str){
    for (var i=0;i<str.length;i++){
         smallarr.push(str)
      }

     for (var j=0;j<smallarr.length;j++){
       smallarr[j] = smallarr[j].slice(0,j) + smallarr[j][j].toUpperCase() + smallarr[j].slice(j+1);
     }
   return smallarr     
   }

  document.write(wave("coding"))
  console.log(smallarr)

Alternatively, achieve the same result with just one loop:

   function wave(str){
    var smallarr=[]
    for (var i=0;i<str.length;i++){
         smallarr.push(str.slice(0,i) + str[i].toUpperCase() + str.slice(i+1))
      }
     return smallarr     
   }
  console.log(wave("coding"))

Answer 6

If you want to experiment with Array.map(), consider using the second callback parameter to represent an index:

let input = "coding";

let result = input.split('').map(
     (_, i) => input.split('').map(
          (char, j) => i === j ? char.toUpperCase() : char).join()
 );

console.log(result);

UPDATE:

An issue with your method is that strings are immutable, so a new string should be constructed using slice as shown below:

var smallarr=[]
   var bigarr=[]
   var another=""
   function wave(str){
    for (var i=0;i<str.length;i++){
         smallarr.push(str)
      }

     for (var j=0;j<smallarr.length;j++){
       smallarr[j] = smallarr[j].slice(0,j) + smallarr[j][j].toUpperCase() + smallarr[j].slice(j+1);
     }
   return smallarr     
   }

  document.write(wave("coding"))
  console.log(smallarr)

Alternatively, achieve the same result with just one loop:

   function wave(str){
    var smallarr=[]
    for (var i=0;i<str.length;i++){
         smallarr.push(str.slice(0,i) + str[i].toUpperCase() + str.slice(i+1))
      }
     return smallarr     
   }
  console.log(wave("coding"))

Answer 7

Answer №4

Your code faces a key issue where you attempt to update a character within a string, which is not feasible due to the immutability of strings in JavaScript. Assigning a value to a specific index will not alter the original string.

The recommended approach is to reconstruct a new string with the desired modifications. You can utilize the slice method to extract portions of the string and construct the updated version.

Below is your code modified to address this limitation:

function transformString(str) {
    var newArr = []; // declare as local variable
    for (var i = 0; i < str.length; i++) {
        let char = str[i]; // Introduce a variable to prevent redundancy
        if (char.toUpperCase() === char) {
            char = char.toLowerCase();
        } else {
            char = char.toUpperCase()
        }
        // String manipulation directly is restricted; necessitating generation of a new string
        newArr.push(str.slice(0, i) + char + str.slice(i+1));
    }
    return newArr;
}

console.log(transformString("example"));

Answer 8

Your code faces a key issue where you attempt to update a character within a string, which is not feasible due to the immutability of strings in JavaScript. Assigning a value to a specific index will not alter the original string.

The recommended approach is to reconstruct a new string with the desired modifications. You can utilize the slice method to extract portions of the string and construct the updated version.

Below is your code modified to address this limitation:

function transformString(str) {
    var newArr = []; // declare as local variable
    for (var i = 0; i < str.length; i++) {
        let char = str[i]; // Introduce a variable to prevent redundancy
        if (char.toUpperCase() === char) {
            char = char.toLowerCase();
        } else {
            char = char.toUpperCase()
        }
        // String manipulation directly is restricted; necessitating generation of a new string
        newArr.push(str.slice(0, i) + char + str.slice(i+1));
    }
    return newArr;
}

console.log(transformString("example"));

Answer 9

Answer №5

Function.prototype.swapCase=function(index, char) {
    var arr = this.split("");
    arr[index] = char;
    return arr.join("");
}

const modifyString = ( text ) => {
  console.log('text ' + text.split(''));
  let array = text.split('')
  array.map((letter, i) => {
      let modified = text.swapCase(i, letter.toUpperCase());
      console.log(modified);
      return modified;
    });
    console.log(array)
    
   // console.log(text)
}

modifyString('example');

Answer 10

Function.prototype.swapCase=function(index, char) {
    var arr = this.split("");
    arr[index] = char;
    return arr.join("");
}

const modifyString = ( text ) => {
  console.log('text ' + text.split(''));
  let array = text.split('')
  array.map((letter, i) => {
      let modified = text.swapCase(i, letter.toUpperCase());
      console.log(modified);
      return modified;
    });
    console.log(array)
    
   // console.log(text)
}

modifyString('example');

Answer 11

Answer №6

One way to manipulate a string is to split it, map the array and then check the index before applying an upper case letter to matching indices.

const
    modifyString = (str) => str
        .split('')
        .map((_, i, splitted) => splitted.map((char, j) => i === j ? char.toUpperCase() : char).join(''));
    };

console.log(modifyString('edabit'));

An alternative method

function transformString(string) {
    var result = [],
        i, j, temp;

    for (i = 0; i < string.length; i++) {
        temp = '';
        for (j = 0; j < string.length; j++) {
            temp += i === j ? string[j].toUpperCase() : string[j];
        }
        result.push(temp);
    }

    return result;
}

console.log(transformString('edabit'));

Answer 12

One way to manipulate a string is to split it, map the array and then check the index before applying an upper case letter to matching indices.

const
    modifyString = (str) => str
        .split('')
        .map((_, i, splitted) => splitted.map((char, j) => i === j ? char.toUpperCase() : char).join(''));
    };

console.log(modifyString('edabit'));

An alternative method

function transformString(string) {
    var result = [],
        i, j, temp;

    for (i = 0; i < string.length; i++) {
        temp = '';
        for (j = 0; j < string.length; j++) {
            temp += i === j ? string[j].toUpperCase() : string[j];
        }
        result.push(temp);
    }

    return result;
}

console.log(transformString('edabit'));

Answer 13

Answer №7

For smaller arrays, an efficient method involves utilizing the split, map, and join functions as shown below:

let word = `edabit`;

let output = word.split('').map((char,index,arr)=>{
  let tempArr = [...arr];
  tempArr[index]=tempArr[index].toUpperCase();
  return tempArr.join('');
});

console.log(output);

However, this approach is not optimized for larger arrays.

Answer 14

For smaller arrays, an efficient method involves utilizing the split, map, and join functions as shown below:

let word = `edabit`;

let output = word.split('').map((char,index,arr)=>{
  let tempArr = [...arr];
  tempArr[index]=tempArr[index].toUpperCase();
  return tempArr.join('');
});

console.log(output);

However, this approach is not optimized for larger arrays.

Transform all characters in a string to uppercase sequentially and then save them in an array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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