FrogRiverOne JavaScript solution 100%/100%

frogriverone.js

function solution(X, A) {
  if (A.length === 1) { // If the array is one element
    // And if its first item is 1 as well as the value to search for,
    // the frog doesn't need to move
    if (A[0] === 1 && X === 1) return 0; 
    // If not, it's impossible to go anywhere
    else return -1;
  }
    
  var i = -1, // Counter
    sum = 0, // Comparison sum
    Y = (X * (X+1)) / 2,  // Sum of 1..X
    f = []; // Found elements
    
  do { // Start searching
    i++; // Increase the counter
    // If we've already found the element, continue
    if (f[A[i]]) continue;
    // If we haven't found the element, mark it
    f[A[i]] = true;
    // Add to the comparison sum that we will be using
    // to determine whether the frog will have been
    // able to cross successfully by this point and then
    // compare it
    sum += A[i];
    if (sum === Y) break;
  } while (i < A.length) // If the counter is over the length, we didn't find it
  
  // If we reached this point and this conditional is true,
  // we didn't find the number.
  if (i === A.length) return -1;
    
  return i; // Return how long it took
}

function solution(X, A) {
  let sum = (X * (X + 1)) / 2;
  const hash = {};

  for (let i = 0; i < A.length; i++) {
    const element = A[i];

    if (hash[element]) continue;
    else {
      hash[element] = true;
      sum -= element;

      if (sum === 0) return i;
    }
  }

  return -1;
}

Here's mine.

function solution(target, array) {
    const positionDict = new Set();
    let minSec = 0;

    for (let second=0; second <= array.length - 1; second ++){
        const position = array[second];

        if (positionDict.has(position)){
            continue;
        }

        minSec = Math.max(minSec, second);
        positionDict.add(position)
    }

    return positionDict.size === target ? minSec : -1;

}

Here's my solution, using the findIndex method which already satisfies the -1 request when no solution is found.

function solution(X, A){
  const set = new Set()
  return A.findIndex((el, i) => {
    set.add(el)
    return set.size === X
  })
}