How can you identify the second least common integer in an array of numbers?

Question

How can you identify the second least common integer in an array of numbers?

Seeking help with a JavaScript coding challenge that I'm stuck on, here is the question:

Create a function that takes an array of integers and returns an element from that array.

The function should calculate the frequency of each element (how many times it appears in the array) and ideally return the element with the second-lowest frequency. If that's not possible, it should return the element with the lowest frequency.

If there are multiple elements that meet the criteria, the second smallest one (based on value) should be returned.

Examples:

secondLowest( [4, 3, 1, 1, 2] ) === 1

secondLowest( [4, 3, 1, 1, 2, 2] ) === 2

secondLowest( [4, 3, 1, 2] ) === 2

Here is what I have so far, but I'm unsure how to proceed:


    function mode(array) {
  if (array.length == 0) return null;
  let modeMap = {};
  let maxEl = array[0],
    maxCount = 1;
  for (let i = 0; i < array.length; i++) {
    var el = array[i];
    if (modeMap[el] == null) modeMap[el] = 1;
    else modeMap[el]++;
    if (modeMap[el] > maxCount) {
      maxEl = el;
      maxCount = modeMap[el];
    }
  }
  return maxEl;
}

javascript arrays frequency

Answer 1

Answer №1

I made it a point to provide a flexible, parameterized function that doesn't rely on hardcoded numbers.

Your instance included two fixed values:

The second least frequency should be chosen
In cases of ties, the second least value should be chosen

Here's how the code operates:

Determine the frequency of each input value
Group together values with the same frequency
Sort these grouped pairs by frequency and select the nth-lowest (in this case, n=2)
If there are multiple pairs with the nth-lowest frequency, sort them by value, and choose the mth-lowest pair (in your example, m=2)
Return the value of this final pair

The m and n parameters mentioned here are denoted as freqInd and valInd in the code. Keep in mind that to choose the second-lowest frequency, freqInd should be 1, not 2 (as 0 would select the lowest, hence 1 picks the second-lowest).

let lowestFreqVal = (freqInd, valInd, values) => {
  
  // Compute frequencies in a map
  let f = new Map();
  for (let v of values) f.set(v, (f.get(v) || 0) + 1);
  
  // Group together all val/freq pairs with the same frequency
  let ff = new Map();
  for (let [ val, freq ] of f) ff.set(freq, (ff.get(freq) || []).concat([ val ]));
  
  // Sort these groups by frequency
  let byFreq = [ ...ff ].sort(([ freq1 ], [ freq2 ]) => freq1 - freq2);  
  
  // Here are all the items of the `freqInd`-th lowest frequency, sorted by value
  // Note that `[1]` returns an array of integers at the frequency, whereas `[0]` would return the frequency itself
  let lowestItems = byFreq[ Math.min(byFreq.length - 1, freqInd) ][1]
    .sort((v1, v2) => v1 - v2);
  
  // Return the `valInd`-th lowest value
  return lowestItems[ Math.min(lowestItems.length - 1, valInd) ];
  
};

console.log('Some random examples:');
for (let i = 0; i < 10; i++) {
  // An array of random length, full of random integers
  let arr = [ ...new Array(3 + Math.floor(Math.random() * 5)) ]
    .map(v => Math.floor(Math.random() * 4));
  
  // Show the result of `lowestFreqVal` on this random Array
  console.log(`lowestFreqVal(1, 1, ${JSON.stringify(arr)}) = ${lowestFreqVal(1, 1, arr)}`);
}

This approach may not be optimal, as it relies on using sort. It is recognized that finding the nth-maximal value in a list can be achieved with a faster runtime than sort (especially when n is small - logically, if n=0, a single pass (O(n)) suffices).

Answer 2

I made it a point to provide a flexible, parameterized function that doesn't rely on hardcoded numbers.

Your instance included two fixed values:

The second least frequency should be chosen
In cases of ties, the second least value should be chosen

Here's how the code operates:

Determine the frequency of each input value
Group together values with the same frequency
Sort these grouped pairs by frequency and select the nth-lowest (in this case, n=2)
If there are multiple pairs with the nth-lowest frequency, sort them by value, and choose the mth-lowest pair (in your example, m=2)
Return the value of this final pair

The m and n parameters mentioned here are denoted as freqInd and valInd in the code. Keep in mind that to choose the second-lowest frequency, freqInd should be 1, not 2 (as 0 would select the lowest, hence 1 picks the second-lowest).

let lowestFreqVal = (freqInd, valInd, values) => {
  
  // Compute frequencies in a map
  let f = new Map();
  for (let v of values) f.set(v, (f.get(v) || 0) + 1);
  
  // Group together all val/freq pairs with the same frequency
  let ff = new Map();
  for (let [ val, freq ] of f) ff.set(freq, (ff.get(freq) || []).concat([ val ]));
  
  // Sort these groups by frequency
  let byFreq = [ ...ff ].sort(([ freq1 ], [ freq2 ]) => freq1 - freq2);  
  
  // Here are all the items of the `freqInd`-th lowest frequency, sorted by value
  // Note that `[1]` returns an array of integers at the frequency, whereas `[0]` would return the frequency itself
  let lowestItems = byFreq[ Math.min(byFreq.length - 1, freqInd) ][1]
    .sort((v1, v2) => v1 - v2);
  
  // Return the `valInd`-th lowest value
  return lowestItems[ Math.min(lowestItems.length - 1, valInd) ];
  
};

console.log('Some random examples:');
for (let i = 0; i < 10; i++) {
  // An array of random length, full of random integers
  let arr = [ ...new Array(3 + Math.floor(Math.random() * 5)) ]
    .map(v => Math.floor(Math.random() * 4));
  
  // Show the result of `lowestFreqVal` on this random Array
  console.log(`lowestFreqVal(1, 1, ${JSON.stringify(arr)}) = ${lowestFreqVal(1, 1, arr)}`);
}

This approach may not be optimal, as it relies on using sort. It is recognized that finding the nth-maximal value in a list can be achieved with a faster runtime than sort (especially when n is small - logically, if n=0, a single pass (O(n)) suffices).

How can you identify the second least common integer in an array of numbers?

Answer №1

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