Aggressive Cows Maximum Minimum Distance in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the stalls array.

DSA Typescript

stalls.sort((a, b) => a [1] b);

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A+

B>

C<

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '-' causes descending order.

Using '-' directly without a function returns NaN.

The sort function needs a comparison that returns a negative value when a is less than b, so using '-' in the comparison function is correct.

2fill in blank

medium

Complete the code to check if cows can be placed with at least mid distance.

DSA Typescript

if (stalls[i] - lastPos [1] mid) {

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A<

B>=

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' places cows too close.

Using '==' is too strict and misses valid placements.

We want to place a cow only if the distance between current stall and last placed cow is at least mid, so '>=' is correct.

3fill in blank

hard

Fix the error in the binary search loop condition.

DSA Typescript

while (low [1] high) {

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A<=

B<

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' misses checking when low equals high.

Using '>' or '>=' reverses the logic.

The binary search continues while low is less than or equal to high to cover all possible distances.

4fill in blank

hard

Fill both blanks to update the binary search bounds correctly.

DSA Typescript

if (canPlaceCows(stalls, cows, mid)) {
  result = mid;
  low = mid [1] 1;
} else {
  high = mid [2] 1;
}

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A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping '+' and '-' causes infinite loops.

Using '*' or '/' is incorrect for adjusting bounds.

If cows can be placed, we try for a bigger distance by increasing low. Otherwise, we decrease high to try smaller distances.

5fill in blank

hard

Fill the blanks to complete the canPlaceCows function logic.

DSA Typescript

function canPlaceCows(stalls: number[], cows: number, mid: number): boolean {
  let count = 1;
  let lastPos = stalls[0];
  for (let i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos [1] mid) {
      count++;
      lastPos = stalls[i];
      if (count [2] cows) {
        return true;
      }
    }
  }
  return false;
}

Drag options to blanks, or click blank then click option'

A>=

B==

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' for distance check is too strict.

Using '<=' for count comparison is incorrect.

We check if distance is at least mid (>=), increment count, and if count is greater than or equal to cows, return true. The lastPos update is already in code.