HackerRank Challenge: Strategies for Efficiently Solving Minimum Swaps 2

In this challenge, the goal is to determine the minimum number of swaps needed to arrange an array of disordered consecutive digits in ascending order. My code successfully handles most of the tests, but I'm encountering timeout errors with four specific cases. Can anyone help pinpoint why my code is timing out? Is there a more efficient approach that could provide quicker results?

function minimumSwaps(arr) {
const min = Math.min(...arr);
let swapCount = 0;
const swap = (array, a, b) => {
    let temp = array[a];
    array[a] = array[b];
    array[b] = temp;
}
for(let i=0; i<arr.length; i++){
    if(arr[i]!==i+min){
        swap(arr, i, arr.indexOf(i+min));
        swapCount++;
    }
}
return swapCount;
}

I initially believed my solution was efficient as it only requires a single iteration over the array's length. I am eager to gain insights into why this implementation isn't meeting performance expectations.

Answer №1

Your main concern arises from the usage of arr.indexOf, which necessitates scanning the entire array every time a swap occurs. An effective solution is to create a reverse lookup mapping values to their corresponding indices before commencing the sorting process, and then updating this list throughout the sort. It's important to note that full swaps are unnecessary; you only need to copy the value from arr[i] to its rightful position since each number is visited just once. Additionally, there's no requirement for determining the minimum value (min) as it’s always guaranteed to be 1 according to the specified conditions, and the last element in the array doesn't require inspection as it must already be correctly positioned by the time it's reached.

function minimumSwaps(arr) {
  const indexes = arr.reduce((c, v, i) => (c[v] = i, c), []);
  const len = arr.length - 1;
  let swapCount = 0;
  for (let i = 0; i < len; i++) {
    if (arr[i] !== i + 1) {
      arr[indexes[i+1]] = arr[i];
      indexes[arr[i]] = indexes[i+1];
      swapCount++;
    }
  }
  return swapCount;
}

console.log(minimumSwaps([7, 1, 3, 2, 4, 5, 6]));
console.log(minimumSwaps([4, 3, 1, 2]));
console.log(minimumSwaps([2, 3, 4, 1, 5]));
console.log(minimumSwaps([1, 3, 5, 2, 4, 6, 7]));

Answer №2

I believe it would be more beneficial to reverse the function rather than search through it.

function minimumSwaps($arr) {
$outp = 0;
$result = array_flip($arr);
for($i=0; $i<count($arr); $i++){
   if($arr[$i] != $i +1){
        $keyswp = $result[$i + 1];
        $temp = $arr[$i];
        $arr[$i] = $i + 1;
        $arr[$keyswp] = $temp; 
        $tempr = $result[$i + 1];
        $result[$i + 1] = $i +1;
        $result[$temp] = $keyswp;
        $outp = $outp + 1;
    } 
}
return $outp;     

}

Answer №3

Presented here is my unique solution, which takes inspiration from Nick's approach but with a twist. Rather than utilizing the Array.Prototype.indexOf method, I opted for an object literal to map the indices of each value. This choice was made to enhance Time complexity since searching in an object literal operates at O(1) (One could also consider using ES6 maps). Here is the implementation:

function minimumSwaps(arr) {
    
    let count = 0;
    let search = arr.reduce((o, e, i) => {
        o[e] = i
        return o
    }, {})
    
    

    for(let i = 0 ; i < arr.length - 1; i++){
        let j = search[i + 1]
        
        if(arr[i] !== i + 1) {
            [arr[i], arr[j]] = [arr[j], arr[i]]
            
            search[arr[i]] = i;
            search[arr[j]] = j;
            count += 1
        }
        
    }
    
    console.log(arr)
    return count
} 

Answer №4

It has been recommended by others that you avoid using the indexOf method because it can make your solution inefficient with a time complexity of O(n^2). This is due to the fact that the indexOf method needs to scan the entire array each time it is called. A better approach would be to create a position map array beforehand and utilize it during the swap process. This will ensure that your solution remains linear in terms of complexity. For further details and solutions to the HackerRank Minimum Swaps 2 Problem in languages such as java, c, c++, and js, you can refer to thisdetailed explanation and solution.

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