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DSA C++programming~10 mins

Find Maximum Subarray Divide and Conquer in DSA C++ - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to find the middle index for dividing the array.

DSA C++
int mid = (left [1] right) / 2;
Drag options to blanks, or click blank then click option'
A*
B-
C+
D/
Attempts:
3 left
💡 Hint
Common Mistakes
Using subtraction instead of addition
Using multiplication instead of addition
2fill in blank
medium

Complete the code to recursively find the maximum subarray sum in the left half.

DSA C++
int left_sum = findMaxSubarray(arr, left, [1]);
Drag options to blanks, or click blank then click option'
Amid
Bleft
Cright
Dmid + 1
Attempts:
3 left
💡 Hint
Common Mistakes
Using right instead of mid
Using mid + 1 instead of mid
3fill in blank
hard

Fix the error in the loop that finds the maximum sum crossing the middle from left to right.

DSA C++
for (int i = mid; i >= left; i--) {
    sum += arr[i];
    if (sum [1] left_sum) {
        left_sum = sum;
    }
}
Drag options to blanks, or click blank then click option'
A<
B>=
C<=
D>
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<' instead of '>'
Using '<=' or '>=' which may cause incorrect updates
4fill in blank
hard

Fill the blanks to find the maximum subarray sum crossing the middle index.

DSA C++
int max_crossing_sum = left_sum [1] right_sum;
return [2](max_crossing_sum, [3](left_max, right_max));
Drag options to blanks, or click blank then click option'
A+
Bmax
C,
D-
Attempts:
3 left
💡 Hint
Common Mistakes
Using subtraction instead of addition
Using commas instead of operators
Not using max function correctly
5fill in blank
hard

Fill all three blanks to complete the recursive function call and return statement for maximum subarray sum.

DSA C++
int left_max = findMaxSubarray(arr, left, [1]);
int right_max = findMaxSubarray(arr, [2], right);
return [3](left_max, right_max, max_crossing_sum);
Drag options to blanks, or click blank then click option'
Amid
Bmid + 1
Cmax
Dmin
Attempts:
3 left
💡 Hint
Common Mistakes
Using min instead of max
Swapping mid and mid + 1 in recursive calls

Practice

(1/5)
1. What is the main idea behind the Divide and Conquer approach to find the maximum subarray sum?
easy
A. Use a sliding window to find the max sum subarray in one pass
B. Sort the array and pick the largest element as max subarray sum
C. Divide the array into two halves, find max subarray in left, right, and crossing middle, then combine results
D. Calculate all possible subarrays and pick the one with max sum

Solution

  1. Step 1: Understand divide and conquer strategy

    The array is split into two halves recursively to simplify the problem.
  2. Step 2: Combine results from left, right, and crossing subarrays

    We find max subarray sums in left half, right half, and subarrays crossing the middle, then take the maximum.
  3. Final Answer:

    Divide the array into two halves, find max subarray in left, right, and crossing middle, then combine results -> Option C
  4. Quick Check:

    Divide and conquer = split and combine max sums [OK]
Hint: Split array, check left, right, and crossing sums [OK]
Common Mistakes:
  • Thinking sorting helps find max subarray sum
  • Ignoring subarrays crossing the middle
  • Trying to check all subarrays without divide and conquer
2. Which of the following is the correct base case condition in the divide and conquer function for maximum subarray?
easy
A. if (low >= high) return arr[low] + arr[high];
B. if (low > high) return 0;
C. if (low == high) return 0;
D. if (low == high) return arr[low];

Solution

  1. Step 1: Identify base case for single element

    When low equals high, the subarray has one element, so max sum is that element itself.
  2. Step 2: Check options for correct base case

    if (low == high) return arr[low]; returns arr[low] when low == high, which is correct base case.
  3. Final Answer:

    if (low == high) return arr[low]; -> Option D
  4. Quick Check:

    Single element subarray max sum = element value [OK]
Hint: Base case: single element returns that element [OK]
Common Mistakes:
  • Returning 0 for single element subarray
  • Using incorrect comparison operators in base case
  • Returning sum of two elements when only one element left
3. What is the output of the following code snippet using divide and conquer to find max subarray sum?
int arr[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
int n = 9;
int maxCrossingSum(int arr[], int low, int mid, int high) {
    int sum = 0, left_sum = INT_MIN;
    for (int i = mid; i >= low; i--) {
        sum += arr[i];
        if (sum > left_sum) left_sum = sum;
    }
    sum = 0; int right_sum = INT_MIN;
    for (int i = mid + 1; i <= high; i++) {
        sum += arr[i];
        if (sum > right_sum) right_sum = sum;
    }
    return left_sum + right_sum;
}
int maxSubArraySum(int arr[], int low, int high) {
    if (low == high) return arr[low];
    int mid = (low + high) / 2;
    int left_max = maxSubArraySum(arr, low, mid);
    int right_max = maxSubArraySum(arr, mid + 1, high);
    int cross_max = maxCrossingSum(arr, low, mid, high);
    return max(max(left_max, right_max), cross_max);
}
int result = maxSubArraySum(arr, 0, n - 1);
std::cout << result;
medium
A. 6
B. 7
C. 5
D. 4

Solution

  1. Step 1: Identify the maximum subarray

    The array is [-2, 1, -3, 4, -1, 2, 1, -5, 4]. The subarray [4, -1, 2, 1] (indices 3-6) sums to 6, the maximum.
  2. Step 2: Verify code execution

    Top-level split: left_max=4, right_max=4, cross_max=6 (left suffix 4 + right prefix -1+2+1=2). Returns 6.
  3. Final Answer:

    6 -> Option A
  4. Quick Check:

    Max subarray sum = 6 [OK]
Hint: Check sums of subarrays crossing middle carefully [OK]
Common Mistakes:
  • Confusing max crossing sum with max overall sum
  • Miscounting indices in loops
  • Assuming max sum includes all positive numbers only
4. Identify the bug in this divide and conquer max subarray code snippet:
int maxCrossingSum(int arr[], int low, int mid, int high) {
    int sum = 0, left_sum = 0;
    for (int i = mid; i >= low; i--) {
        sum += arr[i];
        if (sum > left_sum) left_sum = sum;
    }
    sum = 0; int right_sum = 0;
    for (int i = mid + 1; i <= high; i++) {
        sum += arr[i];
        if (sum > right_sum) right_sum = sum;
    }
    return left_sum + right_sum;
}
medium
A. left_sum and right_sum should be initialized to INT_MIN, not 0
B. The loops should run from low to mid and mid+1 to high respectively
C. sum should not be reset to 0 before right loop
D. The function should return max(left_sum, right_sum) instead of sum

Solution

  1. Step 1: Check initialization of left_sum and right_sum

    They are initialized to 0, which fails if all numbers are negative because max sum could be less than 0.
  2. Step 2: Correct initialization to INT_MIN

    Initializing left_sum and right_sum to INT_MIN ensures negative sums are correctly handled.
  3. Final Answer:

    left_sum and right_sum should be initialized to INT_MIN, not 0 -> Option A
  4. Quick Check:

    Initialize max sums to INT_MIN to handle negatives [OK]
Hint: Use INT_MIN for max sums to handle negative values [OK]
Common Mistakes:
  • Initializing max sums to zero causing wrong results
  • Incorrect loop directions
  • Not resetting sum before right loop
5. Given an array arr = [-1, -2, -3, -4], what will the divide and conquer maximum subarray function return?
hard
A. -10
B. -1
C. 0
D. -4

Solution

  1. Step 1: Understand max subarray with all negative numbers

    When all numbers are negative, max subarray is the single largest element (least negative).
  2. Step 2: Find largest element in arr

    Elements are -1, -2, -3, -4. The largest is -1.
  3. Final Answer:

    -1 -> Option B
  4. Quick Check:

    Max subarray in negatives = max element [OK]
Hint: Max subarray in all negatives is max single element [OK]
Common Mistakes:
  • Returning sum of all elements instead of max single element
  • Returning zero incorrectly assuming empty subarray
  • Ignoring negative values in max calculation
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