Nested List Weight Sum

Question (LC.339)

Given a nested list of integers, return the weighted sum.

Example

I: [[1,1],2,[1,1]]
O: 2*2 + 2 + 2*2 = 10
I: [1,[4,[6]]]
O: 1 + 4 * 2 + 6 * 3 = 1 + 8 + 18 = 27

What about empty list? 0

DFS Approach

1. define subproblem
    dfsSum will sum up all possible sum for the current level
2. recursive calls
    currentSum += sumDfs()
3. stop condition
    end of the list return

Code

public int depthSum2(List<NestedInteger> nestedList) {
    if (nestedList == null || nestedList.size() == 0) {
        return 0;
    }
    int[] sum = new int[1];
    dfsSum(nestedList, sum, 1);
    return sum[0];
}

private void dfsSum(List<NestedInteger> nestedList, 
                    int[] sum, 
                    int level) {
    for (NestedInteger ele : nestedList) {
        if (ele.isInteger()) {
            sum[0] += ele.getInteger() * level;
        } else {
            dfsSum(ele.getList(), sum, level + 1);
        }
    }
}

BFS Approach

What about BFS? It's easy once we see the nested list as tree. When doing the level order traversal, check if it is list, if so put it back in queue.

Code

public int depthSum(List<NestedInteger> nestedList) {
    if (nestedList == null || nestedList.size() == 0) {
        return 0;
    }
    // init queue
    Queue<NestedInteger> queue = new LinkedList<>();
    for (NestedInteger ele : nestedList) {
        queue.offer(ele);
    }
    int level = 1;
    int sum = 0;
    // BFS
    while (!queue.isEmpty()) {
        int levelSize = queue.size();
        for (int i = 0; i < levelSize; i++) {
            NestedInteger current = queue.poll();
            if (current.isInteger()) {
                sum += current.getInteger() * level;
            } else {
                for (NestedInteger ele : current.getList()) {
                    queue.offer(ele);
                }
            }
        }
        level++;
    }
    // return depth sum
    return sum;
}

Follow Up (LC.364)

Now the weight is defined bottom up. The leaf level has weight 1 and so on.

Example

I: [[1,1],2,[1,1]]
O: 8
I: [1,[4,[6]]]
O: 17

DFS Approach

The most obviously approach is doing a two-pass DFS. The first pass is to find the depth. Then the second pass computes the inverse sum with knowledge of the depth.

BFS Approach

Compound sum approach. Layer up as we go.

public int depthSumInverse(List<NestedInteger> nestedList) {
    if (nestedList == null || nestedList.size() == 0) {
        return 0;
    }
    // init queue
    Queue<NestedInteger> queue = new LinkedList<>();
    for (NestedInteger ele : nestedList) {
        queue.offer(ele);
    }
    int totalSum = 0;
    int pastLevels = 0;
    // BFS
    while (!queue.isEmpty()) {
        int levelSize = queue.size();
        int currentLevel = 0;
        for (int i = 0; i < levelSize; i++) {
            NestedInteger current = queue.poll();
            if (current.isInteger()) {
                currentLevel += current.getInteger();
            } else {
                for (NestedInteger ele : current.getList()) {
                    queue.offer(ele);
                }
            }
        }
        // level++;
        pastLevels += currentLevel;
        totalSum += pastLevels;
    }
    // return depth sum
    return totalSum;
}