Check if string contains substring without using indexOf - Javascript

Question

My function is trying to check if string contains substring without use of indexOf or regex match or any standard JS methods.

Please check this jsfiddle: https://jsfiddle.net/09x4Lpj2/

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';
 }else{
 	 contains = 'true';
 } 
 }
 
 console.log(contains);
 }
containsString(string1, string2);

This solution works only when the indexes are the same between the two strings (ex. applegate and apple). But will not work if the indexes are not the same (ex. applegate and gate). How do I manipulate the iterative values correctly so that the function returns true for both situations?

• Ry-: en.wikipedia.org/wiki/String_searching_algorithm
• Bryan Chen: there is not need to convert string to char array. string is already char array
• Nayuki: Two for loops, nested.

Answer 1

you can try this modified script of yours.

var string1 = 'applegate';
var string2 = 'gate';
var string3 = 'apple';
var string4 = 'leg';
var string5 = 'banana';
function containsString(string1, string2){
 var charArray1 = string1.split('');
 var charArray2 = string2.split('');
 var match = 0;
 // iterate from start of 1st string until length of 1st string minus length of 2nd string
 // you don't need to iterate the last part that is not longer than 2nd string since it will be false
 for(var i = 0; i < charArray1.length - charArray2.length + 1; i++){
 // reset match counter on every iteration
 match = 0;
 // iterate the 2nd string
 for(var j = 0; j < charArray2.length; j++){
 // compare corresponding char location
 if(charArray1[i+j] == charArray2[j]){
 match++;
 // just to check in console
 console.log(i, j, match, charArray1[i+j], charArray2[j]);
 } else {
 // just to check in console
 console.log(i, j, match, charArray1[i+j], charArray2[j]);
 // if not match, just skip current check
 break;
 }
 // if match already found, stop the checks, and return true
 if(match == charArray2.length){
 return true;
 }
 }
 }
 // match not found until end of iteration
 return false;
}
console.log(containsString(string1, string2));
console.log(containsString(string1, string3));
console.log(containsString(string1, string4)); 
console.log(containsString(string1, string5)); // haystack does not contain needle
console.log(containsString(string4, string1)); // haystack is shorter than needle

• ameliapond: Thanks you. This is really helpfull. Especially for environments that doesn't support indexOf method like IE < 9 etc...

Answer 2

Welcome to SO. Regex can be used.. Unless even that is also prohibited..

 function containsString(string1, string2){
 console.log(string1.match(string2) != null ? "Yes" : "No");
 }

Regex

• qollers: thank you for the alternative solution :) however, i am trying my best to avoid regex match or any other methods so that I can develop an algorithm for this type of situation.
• Nayuki: This will not work if string2 contains regex metacharacters like .*[]()
• Searching: Ya agreed. But apparently Regex is out of question..So this cannot be THE Answer.. just a starting point. : )

Answer 3

This code has a logical problem,Only to determine whether the last character of A is equal to the corresponding character of B,Maybe the following code is what you want,add a 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);
}

Answer 4

Check this without using any inbuilt functions

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