Number of Nodes With Value One

Number of Nodes With Value One - Problem

There is an undirected connected tree with n nodes labeled from 1 to n and n - 1 edges. You are given the integer n.

The parent node of a node with a label v is the node with the label floor(v / 2). The root of the tree is the node with the label 1.

For example, if n = 7, then the node with the label 3 has the node with the label floor(3 / 2) = 1 as its parent, and the node with the label 7 has the node with the label floor(7 / 2) = 3 as its parent.

You are also given an integer array queries. Initially, every node has a value 0 on it. For each query queries[i], you should flip all values in the subtree of the node with the label queries[i].

Return the total number of nodes with the value 1 after processing all the queries.

Note that:

Flipping the value of a node means that the node with the value 0 becomes 1 and vice versa.
floor(x) is equivalent to rounding x down to the nearest integer.

Input & Output

Example 1 — Basic Tree

$ Input: n = 7, queries = [2,3,1]

› Output: 1

💡 Note: Tree structure: 1 as root, 2,3 as children of 1, 4,5 as children of 2, 6,7 as children of 3. Query 2 flips subtree [2,4,5] → values: [0,1,0,1,1,0,0]. Query 3 flips subtree [3,6,7] → values: [0,1,1,1,1,1,1]. Query 1 flips entire tree → values: [1,0,0,0,0,0,0]. Only node 1 has value 1. Count = 1.

Example 2 — Single Query

$ Input: n = 3, queries = [2]

› Output: 2

💡 Note: Tree: 1 as root, 2,3 as children. Query 2 flips subtree [2]. Final values: node 1=0, 2=1, 3=0. Only nodes 2 has value 1. Count = 1.

Example 3 — Multiple Flips

$ Input: n = 4, queries = [1,1,2]

› Output: 2

💡 Note: Tree: 1 as root, 2,3 as children of 1, 4 as child of 2. Query 1 flips all nodes, query 1 again flips all nodes back, query 2 flips subtree [2,4]. Final values: node 1=0, 2=1, 3=0, 4=1. Count = 2.

Constraints

1 ≤ n ≤ 10⁵
1 ≤ queries.length ≤ 10⁵
1 ≤ queries[i] ≤ n

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is to count how many times each node gets flipped instead of simulating each flip operation. A node is affected by queries on itself and all its ancestors. Best approach counts flips efficiently by traversing up the tree for each node. Time: O(n × log n), Space: O(q)

Common Approaches

✓ Brute Force - Simulate Each Query

⏱️ Time: O(n × q) Space: O(n)

Process each query one by one. For each query node, find all its descendants using the parent-child relationship and flip their values. Count nodes with value 1 at the end.

Optimized - Count Flips per Node

⏱️ Time: O(n × log n + q) Space: O(q)

Instead of simulating each flip, count how many times each node will be flipped by tracking queries that affect it. A node is flipped by all queries on its ancestors (including itself).

Brute Force - Simulate Each Query — Algorithm Steps

Step 1: Initialize all nodes with value 0
Step 2: For each query, find all descendants of the query node
Step 3: Flip the value of each node in the subtree
Step 4: Count nodes with value 1

Visualization

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Step-by-Step Walkthrough

Initial State

All nodes start with value 0

Process Query

Find subtree and flip all nodes

Count Result

Count nodes with value 1

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int n, int* queries, int queriesSize) {
    // Initialize all nodes with value 0
    int* values = (int*)calloc(n + 1, sizeof(int)); // 1-indexed
    
    // Process each query
    for (int i = 0; i < queriesSize; i++) {
        int query = queries[i];
        // Find all nodes in subtree of query node and flip them
        int* stack = (int*)malloc(n * sizeof(int));
        int top = 0;
        stack[top++] = query;
        
        while (top > 0) {
            int node = stack[--top];
            values[node] = 1 - values[node]; // Flip
            
            // Add children (nodes where parent is current node)
            for (int child = 2 * node; child < (n + 1 < 2 * node + 2 ? n + 1 : 2 * node + 2); child++) {
                if (child <= n) {
                    stack[top++] = child;
                }
            }
        }
        free(stack);
    }
    
    // Count nodes with value 1
    int count = 0;
    for (int i = 1; i <= n; i++) {
        count += values[i];
    }
    free(values);
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    int n;
    scanf("%d", &n);
    
    char queriesStr[1000];
    scanf("%s", queriesStr);
    
    // Parse array manually
    int queries[100];
    int queriesSize = 0;
    char* token = strtok(queriesStr + 1, ",]"); // Skip '['
    while (token != NULL) {
        queries[queriesSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(n, queries, queriesSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × q)

For each query, we might visit all n nodes to find descendants

✓ Linear Growth

Space Complexity

O(n)

Array to store node values

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Simulate Each Query — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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