Create a function that accepts a string and groups repeated values

Question

Create a function that accepts a string and groups repeated values. The groups should have the following structure: [[value, first_index, last_index, times_repeated], ..., [value, first_index, last_index, times_repeated]].

value: Character being assessed.
first_index: Index of characters first appearance.
last_index: Index of characters last appearance.
times_repeated: Number of consecutive times character repeats.

Examples

findRepeating("a") ➞ [["a", 0, 0, 1]]

findRepeating("aabbb") ➞ [["a", 0, 1, 2], ["b", 2, 4, 3]]

findRepeating("1337") ➞ [["1", 0, 0, 1], ["3", 1, 2, 2], ["7", 3, 3, 1]]

findRepeating("aabbbaabbb") ➞ [["a", 0, 1, 2], ["b", 2, 4, 3], ["a", 5, 6, 2], ["b", 7, 9, 3]]

I am able to do it for unique characters. But unable to do it for

Number of consecutive times character repeats

MY CODE

function findRepeating(str) {
    let unique = [...new Set([...str])]
    return unique.map(x=>[x,str.indexOf(x),str.lastIndexOf(x),[...str].filter(a=>a==x).length])
}

EXPECTED RESULT

Test.assertSimilar(findRepeating(''), [])

Test.assertSimilar(findRepeating('a'), [['a', 0, 0, 1]])

Test.assertSimilar(findRepeating('1337'), [['1', 0, 0, 1], ['3', 1, 2, 2], ['7', 3, 3, 1]])

Test.assertSimilar(findRepeating('aabbb'), [['a', 0, 1, 2], ['b', 2, 4, 3]])

Test.assertSimilar(findRepeating('addressee'), [['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 4, 1], ['s', 5, 6, 2], ['e', 7, 8, 2]])

Test.assertSimilar(findRepeating('aabbbaabbb'), [['a', 0, 1, 2], ['b', 2, 4, 3], ['a', 5, 6, 2], ['b', 7, 9, 3]])

Test.assertSimilar(findRepeating('1111222233334444'), [['1', 0, 3, 4], ['2', 4, 7, 4], ['3', 8, 11, 4], ['4', 12, 15, 4]])

Test.assertSimilar(findRepeating('1000000000000066600000000000001'), [['1', 0, 0, 1], ['0', 1, 13, 13], ['6', 14, 16, 3], ['0', 17, 29, 13], ['1', 30, 30, 1]])

ACTUAL RESULT

Test Passed: Value == '[]'

Test Passed: Value == "[['a', 0, 0, 1]]"

Test Passed: Value == "[['1', 0, 0, 1], ['3', 1, 2, 2], ['7', 3, 3, 1]]"

Test Passed: Value == "[['a', 0, 1, 2], ['b', 2, 4, 3]]"

FAILED: Expected: "[['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 4, 1], ['s', 5, 6, 2], ['e', 7, 8, 2]]", instead got: "[['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 8, 3], ['s', 5, 6, 2]]"

FAILED: Expected: "[['a', 0, 1, 2], ['b', 2, 4, 3], ['a', 5, 6, 2], ['b', 7, 9, 3]]", instead got: "[['a', 0, 6, 4], ['b', 2, 9, 6]]"

Test Passed: Value == "[['1', 0, 3, 4], ['2', 4, 7, 4], ['3', 8, 11, 4], ['4', 12, 15, 4]]"

FAILED: Expected: "[['1', 0, 0, 1], ['0', 1, 13, 13], ['6', 14, 16, 3], ['0', 17, 29, 13], ['1', 30, 30, 1]]", instead got: "[['1', 0, 30, 2], ['0', 1, 29, 26], ['6', 14, 16, 3]]"

function findRepeating(str) {
	let unique = [...new Set([...str])]
	return unique.map(x=>[x,str.indexOf(x),str.lastIndexOf(x),[...str].filter(a=>a==x).length])
}
console.log("Fails   ",JSON.stringify(findRepeating('addressee')),"\nexpected", `[['a',0,0,1],['d',1,2,2],['r',3,3,1],['e',4,4,1],['s',5,6,2],['e',7,8,2]]`)
console.log("Fails   ",JSON.stringify(findRepeating('aabbbaabbb')),"\nexpected", `[['a',0,1,2],['b',2,4,3],['a',5,6,2],['b',7,9,3]]`)
console.log("Passes  ",JSON.stringify(findRepeating('1111222233334444')),"\nexpected", `[['1',0,3,4],['2',4,7,4],['3',8,11,4],['4',12,15,4]]`)
console.log("Fails   ",JSON.stringify(findRepeating('1000000000000066600000000000001')),"\nexpected", `[['1',0,0,1],['0',1,13,13],['6', 14,16,3],['0',17,29,13],['1',30,30,1]]`)

Nina Scholz · Accepted Answer · 2019-11-07 08:34:39Z

1

You could take an array of same characters with a regular expression which looks for a character and same following onces as a group and map the wanted information.

function findRepeating(string) {
    var i = -1;
    return (string.match(/(.)\1*/g) || []).map(s => [s[0], ++i, i += s.length - 1, s.length]);
}

console.log(findRepeating(""));           // []
console.log(findRepeating("a"));          // [["a", 0, 0, 1]]
console.log(findRepeating("aabbb"));      // [["a", 0, 1, 2], ["b", 2, 4, 3]]
console.log(findRepeating("1337"));       // [["1", 0, 0, 1], ["3", 1, 2, 2], ["7", 3, 3, 1]]
console.log(findRepeating("aabbbaabbb")); // [["a", 0, 1, 2], ["b", 2, 4, 3], ["a", 5, 6, 2], ["b", 7, 9, 3]]

answered Nov 7, 2019 at 8:34

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Nina Scholz Over a year ago

@AshishKamble, which part do you not understand?

Ashish Kamble Over a year ago

After while, I understood, it takes some time for me to understand your answer logic, Awsome, you can put some more words supporting great code

trincot · Accepted Answer · 2019-11-07 08:34:44Z

Your code assumes that each group is about a different character. As soon as you have two groups with the same character, things like lastIndex will give the wrong result.

Just use a plain for-loop.

function findRepeating(str) {
    let result = [];
    let start = 0;
    for (let i = 0; i < str.length; i++) {
        if (str[i] !== str[i+1]) {
            result.push([str[i], start, i, i-start+1]);
            start = i+1;
        }
    }
    return result;
}

console.log(findRepeating(''), [])
console.log(findRepeating('a'), [['a', 0, 0, 1]])
console.log(findRepeating('1337'), [['1', 0, 0, 1], ['3', 1, 2, 2], ['7', 3, 3, 1]])
console.log(findRepeating('aabbb'), [['a', 0, 1, 2], ['b', 2, 4, 3]])
console.log(findRepeating('addressee'), [['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 4, 1], ['s', 5, 6, 2], ['e', 7, 8, 2]])

Collectives™ on Stack Overflow

Create a function that accepts a string and groups repeated values

Examples

MY CODE

EXPECTED RESULT

ACTUAL RESULT

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Examples

MY CODE

EXPECTED RESULT

ACTUAL RESULT

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related