Sherlock 和 Valid String 黑客挑战

Question

I tried to solve this specific challenge in Hackerank and I can't pass all the test cases. 2 is the minimum number of removals required to make it a valid string. It can be done in following two ways:

input = aabbcd...

Sample Output must be no....

Explanation

Remove c and d to get aabb. Or remove a and b to get abcd.

here is my code:

function processData(input) {
    //Enter your code here
   var unique = "";
    var counter = 0;
    var different ="";
  for (i = 0; i < input.length; i++) {
    if (input.lastIndexOf(input[i]) == input.indexOf(input[i])) {
      unique += input[i];

    }counter ++;
  var count1 = (input.match(/input[i]/) || []).length;
  }
    if (counter <= 1)
    {
        console.log("YES");
    }
    else if (counter > 1){
        console.log("NO"); 
    }
    //console.log(count1);
    //console.log(unique);
}

Answer 1

Your algorithms is failing because your solution is not solving the problem. Your final check to see whether the String is acceptable or not uses counter

  if (counter <= 1)
    {
        console.log("YES");
    }
    else if (counter > 1){
        console.log("NO"); 
    }

However counter is incremented everytime your for loop runs. This means your solutions is only passing the tests that resolve to "NO" by default.

Also your created a variable called different which you never used and both unique and count1 are assigned values but never used afterwards.

Your algorithm should check the frequency of each string and see if they are the same (or a most different by 1) for each string.Your algorithm only checks for uniqueness of a letter. I would suggest counting the occurence ofeach letter using

input.match(/input[i]/).length;

and the comparing the frequencies

Answer 2

Here is my answer. Points to remember 1. Make sure you trim else your get escape \\n as part of input. 2. Count occurrences of chars. 3. map char count to their occurrence. 4. Now check if occurrence is 1 => valid else if 2 make sure you check the at least one occurrence count is 1 ex: aabbcd c and d occurs 2 times and both have char count as 1 so he has to remove both c and d.

Note: Not clarified in problem same char can be removed more than once and it still can be considered as one removal ex: aabbcccc I have 4 c's but I can use one operation to remove 2 C's to make it valid string aabbcc.

var bIsValidString = false;
var oCharMapCount = input.trim().split("").reduce((acc, val) => {
    if (acc[val]) {
        acc[val] += 1;
    }
    else {
        acc[val] = 1;
    }
    return acc;
}, {});
var oCharOccuranceMap = {}; 
for(let key in oCharMapCount) {
    if (oCharOccuranceMap[oCharMapCount[key]]) {
        oCharOccuranceMap[oCharMapCount[key]] += 1;
    }
    else {
        oCharOccuranceMap[oCharMapCount[key]] = 1;
    }
}
var iCharOccurances = (Object.keys(oCharOccuranceMap) || []).length;

if (iCharOccurances === 1) {
    bIsValidString = true;
}
else if (iCharOccurances === 2) {
    for (let iCharCountOccurance in oCharOccuranceMap) {
        if (oCharOccuranceMap[iCharCountOccurance] === 1) {
            bIsValidString = true;
        }
    }
}
else {
    bIsValidString = false;
}

(bIsValidString) ? console.log("YES") : console.log("NO");

Answer 3

After a few iterations with 2 test cases failing, i finally came up with the correct solution below.

//create histogram to count occurances of each char in string
let chars               = s.reduce(into: [:]) { $0[$1, default:0] += 1 }

//create set for occurances to determine the number of different occurances found
let valueSet:Set<Int> = Set(chars.values)
var isValid = false

//if all chars have same occurrance
if valueSet.count == 1 {
    isValid = true
}
//if there are more than two different occurance counts
else if valueSet.count > 2 {
    isValid = false
} 
//check different frequencies counts
else {
    
    let freq1 = valueSet.min()!
    let freq2 = valueSet.max()!
    
    var freq1Count = 0
    var freq2Count = 0
    
    //loop through all values to count the amount of times each different occurance count occur
    for value in chars.values {
        switch value {
        case freq1:
            freq1Count += 1
        default:
            freq2Count += 1
        }
    }
    
    //if either freq count is 1 and it that count only occurs once
    if (freq1 == 1 && freq1Count == 1) ||
       (freq2 == 1 && freq2Count == 1) {
        isValid = true
    }
    //if the difference between frequencies is one and there is only one of them
    else if abs(freq1 - freq2) == 1 && (freq1Count == 1 || freq2Count == 1) {
        isValid = true
    }
    else {
        isValid = false
    }
}

return isValid ? "YES" : "NO"