leetcode

Distribute Elements Into Two Arrays I - Link

Question Description

You are given an integer array nums.

You must split it into two arrays list1 and list2 such that:

• list1 starts with nums[0]
• list2 starts with nums[1]
• For every i from 2 to n-1, compare the last elements of list1 and list2:
  • If last element of list1 > last element of list2, put nums[i] in list1
  • Otherwise, put nums[i] in list2

Finally, return the concatenation of list1 followed by list2.

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Constraints

• 2 <= nums.length <= 100
• 1 <= nums[i] <= 100

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Approach

Use two ArrayLists to simulate the distribution process with comparison-based placement.

Algorithm:

1. Initialize list1 with nums[0] and list2 with nums[1]
2. For each element from index 2 to n-1:
   • Compare last element of list1 with last element of list2
   • If list1.last > list2.last, add to list1
   • Otherwise, add to list2
3. Create result array and copy elements from list1 followed by list2
4. Return the concatenated result

This approach works because:

• We follow the exact rules specified in the problem
• ArrayLists allow efficient addition to end (O(1) amortized)
• The comparison is always between current last elements
• Final concatenation maintains the required order

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Dry Run

Example 1: nums = [2, 1, 3, 3]

Step-by-step execution:

• Initialize: list1 = [2], list2 = [1]
• i=2, nums[2]=3
  • Compare last of list1 (2) vs last of list2 (1)
  • 2 > 1, so add 3 to list1
  • list1 = [2, 3], list2 = [1]
• i=3, nums[3]=3
  • Compare last of list1 (3) vs last of list2 (1)
  • 3 > 1, so add 3 to list1
  • list1 = [2, 3, 3], list2 = [1]
• Concatenate: [2, 3, 3, 1]

Result: [2, 3, 3, 1]

Example 2: nums = [5, 4, 3, 8]

Step-by-step execution:

• Initialize: list1 = [5], list2 = [4]
• i=2, nums[2]=3
  • Compare last of list1 (5) vs last of list2 (4)
  • 5 > 4, so add 3 to list1
  • list1 = [5, 3], list2 = [4]
• i=3, nums[3]=8
  • Compare last of list1 (3) vs last of list2 (4)
  • 3 < 4, so add 8 to list2
  • list1 = [5, 3], list2 = [4, 8]
• Concatenate: [5, 3, 4, 8]

Result: [5, 3, 4, 8]

Example 3: nums = [1, 3]

Step-by-step execution:

• Only 2 elements, no more to process
• Concatenate: [1, 3]

Result: [1, 3]

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Solution

class Solution {
    public int[] resultArray(int[] nums) {
        int length = nums.length;

        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();

        list1.add(nums[0]);
        list2.add(nums[1]);

        for (int i = 2; i < length; i++) {
            int lastList1 = list1.get(list1.size() - 1);
            int lastList2 = list2.get(list2.size() - 1);

            if (lastList1 > lastList2) {
                list1.add(nums[i]);
            } else {
                list2.add(nums[i]);
            }
        }

        int[] ans = new int[length];
        int index = 0;
        for (int ele : list1) {
            ans[index++] = ele;
        }
        for (int ele : list2) {
            ans[index++] = ele;
        }
        return ans;
    }
}

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Time and Space Complexity

• Time Complexity: O(n) - single pass through nums, linear concatenation
• Space Complexity: O(n) - for two ArrayLists (excluding output array)

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Alternative Approaches

Array-Based Approach (More Memory Efficient)

class Solution {
    public int[] resultArray(int[] nums) {
        int n = nums.length;
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();

        list1.add(nums[0]);
        list2.add(nums[1]);

        for (int i = 2; i < n; i++) {
            if (list1.get(list1.size() - 1) > list2.get(list2.size() - 1)) {
                list1.add(nums[i]);
            } else {
                list2.add(nums[i]);
            }
        }

        // Build result array
        int[] result = new int[n];
        int idx = 0;
        for (int num : list1) result[idx++] = num;
        for (int num : list2) result[idx++] = num;

        return result;
    }
}

Explanation: Same logic but with more explicit variable naming.

Time Complexity: O(n) - same as main approach Space Complexity: O(n) - same as main approach

Pre-sized ArrayList Approach

class Solution {
    public int[] resultArray(int[] nums) {
        int n = nums.length;
        List<Integer> list1 = new ArrayList<>(n);
        List<Integer> list2 = new ArrayList<>(n);

        list1.add(nums[0]);
        list2.add(nums[1]);

        for (int i = 2; i < n; i++) {
            int last1 = list1.get(list1.size() - 1);
            int last2 = list2.get(list2.size() - 1);

            if (last1 > last2) {
                list1.add(nums[i]);
            } else {
                list2.add(nums[i]);
            }
        }

        int[] result = new int[n];
        for (int i = 0; i < list1.size(); i++) {
            result[i] = list1.get(i);
        }
        for (int i = 0; i < list2.size(); i++) {
            result[list1.size() + i] = list2.get(i);
        }

        return result;
    }
}

Explanation: Pre-size ArrayLists and use indexed assignment for result.

Time Complexity: O(n) - same performance Space Complexity: O(n) - same as main approach

Stream API Approach (Java 8+)

class Solution {
    public int[] resultArray(int[] nums) {
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();

        list1.add(nums[0]);
        list2.add(nums[1]);

        for (int i = 2; i < nums.length; i++) {
            int last1 = list1.get(list1.size() - 1);
            int last2 = list2.get(list2.size() - 1);

            if (last1 > last2) {
                list1.add(nums[i]);
            } else {
                list2.add(nums[i]);
            }
        }

        return Stream.concat(list1.stream(), list2.stream())
                    .mapToInt(Integer::intValue)
                    .toArray();
    }
}

Explanation: Use Stream API for concatenation instead of manual copying.

Time Complexity: O(n) - stream operations Space Complexity: O(n) - same as main approach

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Key Insights

Distribution Strategy:

• Elements tend to go to the array with the larger last element
• This creates a natural separation based on relative magnitudes
• The algorithm is greedy and makes locally optimal choices

Edge Cases:

• n = 2: Return [nums[0], nums[1]] (no distribution needed)
• All increasing: Elements go to list1 if each > previous maximum
• All decreasing: Elements go to list2 after first comparison

Performance Characteristics:

• Simple and intuitive algorithm
• No complex data structures needed
• Works well for small to medium arrays (n ≤ 100)
• The distribution creates natural ordering within each list

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