How to determine if a string includes a particular substring in Javascript without using the indexOf method

Question

How to determine if a string includes a particular substring in Javascript without using the indexOf method

Is it possible to determine if a string contains a specific substring without using indexOf, regex match, or any standard JavaScript methods?

Feel free to review this code snippet on jsfiddle: https://jsfiddle.net/09x4Lpj2/

var string1 = 'applegate';
    var string2 = 'gate';

    function checkSubstring(string1, string2){
  var j = 0;
      var k = 0;
      var contains = 'false';
      var charArray1 = string1.split('');
      var charArray2 = string2.split('');
  
      for(var i = 0; i < charArray2.length; i++){
      j = i;
        if(charArray1[j++] != charArray2[k++]){
      contains = 'false';
        }else{
      contains = 'true';
        }   
      }
  
     console.log(contains);
    }
checkSubstring(string1, string2);

This approach works only when the indexes of the characters in both strings align (e.g., comparing "applegate" and "apple"). However, it fails when the indexes are not the same (e.g., comparing "applegate" and "gate"). How can the iteration values be manipulated to ensure accurate results in both scenarios?

javascript arrays

Answer 1

Answer №1

Give this revised script a shot.

var str1 = 'pineapple';
var str2 = 'apple';
var str3 = 'pine';
var str4 = 'ear';

function containsSubstring(str1, str2){
  var chars1 = str1.split('');
  var chars2 = str2.split('');
  var matchCount = 0;

  for(var x = 0; x < chars1.length - chars2.length + 1; x++){
    matchCount = 0;
    for(var y = 0; y < chars2.length; y++){
      if(chars1[x+y] == chars2[y]){
        matchCount++;
        console.log(x, y, matchCount, chars1[x+y], chars2[y]);
      } else {
        console.log(x, y, matchCount, chars1[x+y], chars2[y]);
        break;
      }
      if(matchCount == chars2.length){
        return true;
      }
    }
  }
  return false;
}

console.log(containsSubstring(str1, str2));
console.log(containsSubstring(str1, str3));
console.log(containsSubstring(str1, str4));   
console.log(containsSubstring(str1, str5)); 
console.log(containsSubstring(str4, str1));

Answer 2

Give this revised script a shot.

var str1 = 'pineapple';
var str2 = 'apple';
var str3 = 'pine';
var str4 = 'ear';

function containsSubstring(str1, str2){
  var chars1 = str1.split('');
  var chars2 = str2.split('');
  var matchCount = 0;

  for(var x = 0; x < chars1.length - chars2.length + 1; x++){
    matchCount = 0;
    for(var y = 0; y < chars2.length; y++){
      if(chars1[x+y] == chars2[y]){
        matchCount++;
        console.log(x, y, matchCount, chars1[x+y], chars2[y]);
      } else {
        console.log(x, y, matchCount, chars1[x+y], chars2[y]);
        break;
      }
      if(matchCount == chars2.length){
        return true;
      }
    }
  }
  return false;
}

console.log(containsSubstring(str1, str2));
console.log(containsSubstring(str1, str3));
console.log(containsSubstring(str1, str4));   
console.log(containsSubstring(str1, str5)); 
console.log(containsSubstring(str4, str1));

Answer 3

Answer №2

Greetings from the world of programming. Regular expressions can come in handy, unless they are also not allowed.

function checkSubstring(mainString, subString) {
    console.log(mainString.includes(subString) ? "Found" : "Not Found");
}

Explore Regular Expressions

Answer 4

Greetings from the world of programming. Regular expressions can come in handy, unless they are also not allowed.

function checkSubstring(mainString, subString) {
    console.log(mainString.includes(subString) ? "Found" : "Not Found");
}

Explore Regular Expressions

Answer 5

Answer №3

The existing code has a flaw in its logic – it only checks if the last character of string A is equal to the corresponding character of string B. Perhaps the following revised version of the code is what you are looking for, which includes an additional line of code:

var string1 = 'applegate';
var string2 = 'gate';

function containsString(string1, string2){
  var j = 0;
  var k = 0;
  var contains = 'false';
  var charArray1 = string1.split('');
  var charArray2 = string2.split('');

  for(var i = 0; i < charArray2.length; i++){
    j = i;
    if(charArray1[j++] != charArray2[k++]){
      contains = 'false';
      break;
    }else{
      contains = 'true';
    }   
  }

  console.log(contains);
}

// Additional line of code here

Answer 6

The existing code has a flaw in its logic – it only checks if the last character of string A is equal to the corresponding character of string B. Perhaps the following revised version of the code is what you are looking for, which includes an additional line of code:

var string1 = 'applegate';
var string2 = 'gate';

function containsString(string1, string2){
  var j = 0;
  var k = 0;
  var contains = 'false';
  var charArray1 = string1.split('');
  var charArray2 = string2.split('');

  for(var i = 0; i < charArray2.length; i++){
    j = i;
    if(charArray1[j++] != charArray2[k++]){
      contains = 'false';
      break;
    }else{
      contains = 'true';
    }   
  }

  console.log(contains);
}

// Additional line of code here

Answer 7

Answer №4

Try this solution without relying on any built-in functions

function customSubstringSearch(longString, shortString){
var count = 0;
for(var i = 0; i < longString.length; i++){
  for(var j = 0; j < shortString.length; j++){
    if(shortString[j] != longString[i + j]){
      break;
    }
    if((j+1) == shortString.length){
      count++;
    }
  }
}
return count;
}

Answer 8

Try this solution without relying on any built-in functions

function customSubstringSearch(longString, shortString){
var count = 0;
for(var i = 0; i < longString.length; i++){
  for(var j = 0; j < shortString.length; j++){
    if(shortString[j] != longString[i + j]){
      break;
    }
    if((j+1) == shortString.length){
      count++;
    }
  }
}
return count;
}

How to determine if a string includes a particular substring in Javascript without using the indexOf method

Answer №1

Answer №2

Answer №3

Answer №4

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