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Order & filter an array

Return the top 3 ids of the lowest value given a single clientId

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Instructions

Write a function that receives an input array, a clientId and a number T and returns the top T ids ordered by the lowest value for the given clientId:

Sample input data:

const input = [
    { id: 1, clientId: 12, value: 13 },
    ...
  ];
const clientId = 12;
const T = 3;
Tests
0/5
Function Call:
Expected:
topIds([{ id: 1, clientId: 12, value: 13 }, { id: 2, clientId: 12, value: 10 }, { id: 3, clientId: 12, value: 15 }], 12, 2)
[ 2, 1 ]
Run test
topIds([{ id: 1, clientId: 12, value: 13 }, { id: 2, clientId: 1, value: 10 }, { id: 3, clientId: 12, value: 15 }], 12, 2)
[ 1, 3 ]
Run test
topIds([{ id: 1, clientId: 12, value: 13 }, { id: 2, clientId: 12, value: 15 }, { id: 3, clientId: 12, value: -13 }, { id: 4, clientId: 12, value: 2 }, { id: 5, clientId: 11, value: 11 }, { id: 6, clientId: 10, value: -11 }, { id: 7, clientId: 45, value: -1 }], 12, 3)
[ 3, 4, 1 ]
Run test
topIds([{ id: 1, clientId: 10, value: 13 }], 12, 3)
[]
Run test
topIds([{ id: 1, clientId: 10, value: 13 }], 10, 3)
[ 1 ]
Run test
See Solution

Solution

Basic idea

  1. Filter the input array to include only items with the given clientId.
  2. Sort the filtered array by value in ascending order.
  3. Slice the sorted array to include only the top T items.
  4. Map the resulting array to extract the ids.
const topIds = (input, clientId, T ) => {
    const result = input
    .filter(item => item.clientId === clientId)
    .sort((a, b) => a.value - b.value)
    .slice(0, T)
    .map(a => a.id)

    return result
}

Algorithm Detailed Explanation

  1. Filter the input array to include only items with the given clientId.
  2. Sort the filtered array by value in ascending order.
  3. Slice the sorted array to include only the top T items.
  4. Map the resulting array to extract the ids.

Complexity Analysis

The time complexity of this solution is O(n log n) where n is the number of items in the input array. This is because the sort function has a time complexity of O(n log n). The space complexity is O(n) because we are creating a new array to store the filtered items.

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