Grouping Subarrays in Arrays

Question

I'm wrapping my head around the following logic, but I'm still missing something.

Given an Array like const testArr = ["F", "F", "C", "C", "F", "C", "F"].

The result array should look like ["F", "F", ["C", "C"], "F", ["C"], "F"].

The code I came up until now looks like this:

const grouping = (arr) => {
  const result = [];
  arr.forEach((item, index) => {
    if (item === "C") {
      const subArr = new Array();
      subArr.push(item);
      if (arr[index + 1] !== "C") {
        result.push(subArr);
      }
    } else {
      result.push(item);
    }
  });
  return result;
};

console.log(grouping(testArr));

This prints currently the result:

["F", "F", ["C"], "F", ["C"], "F"]

I appreciate your hints :-)

Answer 1

You could use a while loop with a temporary index to achieve your expected result

Below is the small modifed from your current solution (change to for loop and use while for condition checking)

const testArr = ["F", "F", "C", "C", "F", "C", "F"]

const grouping = (arr) => {
  const result = []
  for (let index = 0; index < arr.length; index++) {
    if (arr[index] === "C") {
      const subArr = [arr[index]]
      let tempIndex = index + 1
      while (arr[tempIndex] === "C") {
        subArr.push(arr[tempIndex])
        index = tempIndex
        tempIndex++
      }
      result.push(subArr)
    } else {
      result.push(arr[index])
    }
  }

  return result
}

console.log(grouping(testArr))

Answer 2

I think I'd do it like this, see comments:

const grouping = arr => {
    const result = [];
    let currentSub = null;
    for (const value of arr) {
        // Is it the special value?
        if (value === "C") {
            // Yes, do we have an active array?
            if (!currentSub) {
                // No, create one and push it
                currentSub = [];
                result.push(currentSub);
            }
            // Add to the active array
            currentSub.push(value)
        } else {
            // Not special, forget active array and push
            currentSub = null;
            result.push(value);
        }
    }
    return result;
};

Live Example:

const testArr = ["F", "F", "C", "C", "F", "C", "F"]

const grouping = arr => {
    const result = [];
    let currentSub = null;
    for (const value of arr) {
        // Is it the special value?
        if (value === "C") {
            // Yes, do we have an active array?
            if (!currentSub) {
                // No, create one and push it
                currentSub = [];
                result.push(currentSub);
            }
            // Add to the active array
            currentSub.push(value)
        } else {
            // Not special, forget active array and push
            currentSub = null;
            result.push(value);
        }
    }
    return result;
};

console.log(grouping(testArr));

.as-console-wrapper {
    max-height: 100% !important;
}

If you prefer a forEach to for-of, it's almost identical:

const testArr = ["F", "F", "C", "C", "F", "C", "F"]

const grouping = arr => {
    const result = [];
    let currentSub = null;
    arr.forEach(value => {
        // Is it the special value?
        if (value === "C") {
            // Yes, do we have an active array?
            if (!currentSub) {
                // No, create one and push it
                currentSub = [];
                result.push(currentSub);
            }
            // Add to the active array
            currentSub.push(value)
        } else {
            // Not special, forget active array and push
            currentSub = null;
            result.push(value);
        }
    });
    return result;
};

console.log(grouping(testArr));

.as-console-wrapper {
    max-height: 100% !important;
}

---

Side note: In general, avoid new Array(). To create a blank array, just use []. To create an array with entries, use [value1, value2] etc. You can use new Array(x) (or just Array(x)) to create a sparse array with the length x, but that's usually only useful when you're about to use fill on it to fill it with the same value in each entry.

Answer 3

I would use a temp array for storing the 'special' value, and I would reset it anytime I encounter a different value.

const grouping = (arr) => {
  const result = []
  let tempC = []
  
  arr.forEach(letter => {
    if (letter === 'C') {
       tempC.push('C')         // Append the special letter to its temp array
    } else {
       if (tempC.length > 0) {
         result.push(tempC)    // If the previus iteration had a 'C', push the array in result
       }
       tempC = []              // Reset the tempC collector
       result.push(letter)     // Add the 'not special' letter to the result
    }
  })
  return result
}

Answer 4

const testArr = ["F", "F", "C", "C", "F", "C", "F"];
const result = testArr.reduce((acc, val) => {
    if (val === "C") {
        Array.isArray(acc[acc.length - 1]) ? acc[acc.length - 1].push(val) : acc.push([val]);
    } else {
        acc.push(val);
    }

    return acc;
}, []);

console.log(result);

Answer 5

This is an approach using the function Array.prototype.reduce which is DRY.

const testArr = ["F", "F", "C", "C", "F", "C", "F"];
const {result} = testArr.reduce((a, e) => {
  if (e === a.target) (a.current || (a.current = [])).push(e);
  else {
    if (a.current) a.result.push(a.current), a.current = undefined;   
    a.result.push(e);
  }
  
  return a;
}, {result: [], current: undefined, target: "C"});

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 6

void groupElements(int arr[], int n) 
{ 
    // Initialize all elements as not visited 
    bool *visited = new bool[n]; 
    for (int i=0; i<n; i++) 
        visited[i] = false; 
  
    // Traverse all elements 
    for (int i=0; i<n; i++) 
    { 
        // Check if this is first occurrence 
        if (!visited[i]) 
        { 
            // If yes, print it and all subsequent occurrences 
            cout << arr[i] << " "; 
            for (int j=i+1; j<n; j++) 
            { 
                if (arr[i] == arr[j]) 
                { 
                    cout << arr[i] << " "; 
                    visited[j] = true; 
                } 
            } 
        } 
    } 
  
    delete [] visited;   
}