Range Sum and Update in Array : Segment Tree using Stack

Given an array arr[] of N integers. The task is to do the following operations:

  1. Add a value X to all the element from from index A to B where 0 ≤ A ≤ B ≤ N-1.
  2. Find the sum of the element from index L to R where 0 ≤ L ≤ R ≤ N-1 before and after the update given to the array above.

Example:

Input: arr[] = {1, 3, 5, 7, 9, 11}, L = 1, R = 3, A = 1, B = 5, X = 10
Output:
Sum of values in given range = 15
Updated sum of values in given range = 45
Explanation:
Sum of values in the range 1 to 3 is 3 + 5 + 7 = 15.
arr[] after adding 10 from index 1 to 5 is arr[] = {1, 13, 15, 17, 19, 21}
Sum of values in the range 1 to 3 after update is 13 + 15 + 17 = 45.



Input: arr[] = { 11, 32, 5, 7, 19, 11, 8}, L = 2, R = 6, A = 1, B = 5, X = 16
Output:
Sum of values in given range = 50
Updated sum of values in given range = 114
Explanation:
Sum of values in the range 2 to 6 is 5 + 7 + 19 + 11 + 8 = 50.
arr[] after adding 16 from index 1 to 5 is arr[] = {11, 48, 21, 23, 35, 27, 8}
Sum of values in the range 2 to 6 after update is 21 + 23 + 35 + 27 + 8 = 114.

Approach:
The recursive approach using a Segment Tree for the given problem is discussed in this article. In this post we will discussed an approach using Stack Data Structure to avoid recursion.

Below are the steps to implement Segment Tree using Stack:

  1. The idea is to use tuple to store the state which has Node number and range indexes in the Stack.
  2. For Building Segment Tree:
    • Push the root Node to the stack as a tuple:
      Stack S;
      start = 0, end = arr_size - 1
      S.emplace(1, start, end)
      
    • Pop the element from the stack and do the following untill stack becomes empty:
      1. If starting index equals to ending index, then we reach the leaf node and update the value at Segment tree array as value at current index in the given array.
      2. Else insert the flag tuple with the current Node as S.emplace(current_node, INF, INF) to inverse the order of evaluation and insert the tuple with value for left and right child as:

        mid = (start + end) / 2
        st.emplace(curr_node * 2, start, mid)
        st.emplace(curr_node * 2 + 1, mid + 1, end)

      3. If start index and end index is same as INF, then update the Segment Tree value at current index as:

        Value at current index is updated as value at left child and right child:
        tree[curr_node] = tree[2*curr_node] + tree[2*curr_node + 1]

  3. For Update Tree:
    • Push the root node to the stack as done for building Segment Tree.
    • Pop the element from the stack and do the following untill stack becomes empty:
      1. If the current node has any pending update then first update to the current node.
      2. If the current node ranges lies completely in the update query range, then update the current node with that value.
      3. If the current node ranges overlap with the update query range, then follow the above approach and push the tuple for left child and right child in the Stack.
      4. Update the query using the result of left and right child above.
  4. For Update Query:
    • Push the root node to the stack as done for building Segment Tree.
    • Pop the element from the stack and do the following untill stack becomes empty:
      1. If the current node ranges lies outside the given query, then continue with the next iteration.
      2. If the current node ranges lies completely in the update query range, then update the result with the current node value.
      3. If the current node ranges overlap with the update query range, then follow the above approach and push the tuple for left child and right child in the Stack.
      4. Update the result using the value obtained from left and right child Node.

Below is the implementation of the above approach:

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#include "bits/stdc++.h"
using namespace std;
  
constexpr static int MAXSIZE = 1000;
constexpr static int INF
    = numeric_limits<int>::max();
  
// Segment Tree array
int64_t tree[MAXSIZE];
  
// Lazy Update array
int64_t lazy[MAXSIZE];
  
// This tuple will hold tree state
// the stacks
using QueryAdaptor
    = tuple<int64_t,
            int64_t,
            int64_t>;
  
// Build our segment tree
void build_tree(int64_t* arr,
                int64_t arr_size)
{
    // Stack will use to update
    // the tree value
    stack<QueryAdaptor> st;
  
    // Emplace the root of the tree
    st.emplace(1, 0, arr_size - 1);
  
    // Repeat until empty
    while (!st.empty()) {
  
        // Take the indexes at the
        // top of the stack
        int64_t currnode, curra, currb;
  
        // value at the top of the
        // stack
        tie(currnode, curra, currb) = st.top();
  
        // Pop the value from the
        // stack
        st.pop();
  
        // Flag with INF ranges are merged
        if (curra == INF && currb == INF) {
            tree[currnode] = tree[currnode * 2]
                             + tree[currnode * 2 + 1];
        }
  
        // Leaf node
        else if (curra == currb) {
            tree[currnode] = arr[curra];
        }
  
        else {
  
            // Insert flag node inverse
            // order of evaluation
            st.emplace(currnode, INF, INF);
  
            int64_t mid = (curra + currb) / 2;
  
            // Push children
            st.emplace(currnode * 2,
                       curra, mid);
            st.emplace(currnode * 2 + 1,
                       mid + 1, currb);
        }
    }
}
  
// A utility function that propagates
// updates lazily down the tree
inline void push_down(int64_t node,
                      int64_t a,
                      int64_t b)
{
    if (lazy[node] != 0) {
        tree[node] += lazy[node] * (b - a + 1);
  
        if (a != b) {
            lazy[2 * node] += lazy[node];
            lazy[2 * node + 1] += lazy[node];
        }
  
        lazy[node] = 0;
    }
}
  
// Iterative Range_Update function to
// add val to all elements in the
// range i-j (inclusive)
void update_tree(int64_t arr_size,
                 int64_t i,
                 int64_t j,
                 int64_t val)
{
  
    // Intialize the stack
    stack<QueryAdaptor> st;
  
    // Emplace the root of the tree
    st.emplace(1, 0, arr_size - 1);
  
    // Work until empty
    while (!st.empty()) {
  
        // Take the indexes at the
        // top of the stack
        int64_t currnode, curra, currb;
        tie(currnode, curra, currb) = st.top();
        st.pop();
  
        // Flag with INF ranges are merged
        if (curra == INF && currb == INF) {
            tree[currnode] = tree[currnode * 2]
                             + tree[currnode * 2 + 1];
        }
  
        // Traverse the previous updates
        // down the tree
        else {
            push_down(currnode, curra, currb);
  
            // No overlap condition
            if (curra > currb || curra > j
                || currb < i) {
                continue;
            }
  
            // Total overlap condition
            else if (curra >= i && currb <= j) {
                // Update lazy array
                tree[currnode] += val * (currb - curra + 1);
  
                if (curra != currb) {
                    lazy[currnode * 2] += val;
                    lazy[currnode * 2 + 1] += val;
                }
            }
  
            // Partial Overlap
            else {
                // Insert flag node inverse
                // order of evaluation
                st.emplace(currnode, INF, INF);
  
                int64_t mid = (curra + currb) / 2;
  
                // Push children
                st.emplace(currnode * 2,
                           curra, mid);
                st.emplace(currnode * 2 + 1,
                           mid + 1, currb);
            }
        }
    }
}
  
// A function that find the sum of
// all elements in the range i-j
int64_t query(int64_t arr_size,
              int64_t i,
              int64_t j)
{
    // Intialize stack
    stack<QueryAdaptor> st;
  
    // Emplace root of the tree
    st.emplace(1, 0, arr_size - 1);
  
    int64_t result = 0;
  
    while (!st.empty()) {
  
        // Take the indexes at the
        // top of the stack
        int64_t currnode, curra, currb;
        tie(currnode, curra, currb) = st.top();
        st.pop();
  
        // Traverse the previous updates
        // down the tree
        push_down(currnode, curra, currb);
  
        // No overlap
        if (curra > currb || curra > j
            || currb < i) {
            continue;
        }
  
        // Total Overlap
        else if (curra >= i && currb <= j) {
            result += tree[currnode];
        }
  
        // Partial Overlap
        else {
            std::int64_t mid = (curra + currb) / 2;
  
            // Push children
            st.emplace(currnode * 2,
                       curra, mid);
            st.emplace(currnode * 2 + 1,
                       mid + 1, currb);
        }
    }
  
    return result;
}
  
// Driver Code
int main()
{
    // Initialize our trees with 0
    memset(tree, 0, sizeof(int64_t) * MAXSIZE);
    memset(lazy, 0, sizeof(int64_t) * MAXSIZE);
  
    int64_t arr[] = { 1, 3, 5, 7, 9, 11 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Build segment tree from given array
    build_tree(arr, n);
  
    // Print sum of values in array
    // from index 1 to 3
    cout << "Sum of values in given range = "
         << query(n, 1, 3)
         << endl;
  
    // Add 10 to all nodes at indexes
    // from 1 to 5
    update_tree(n, 1, 5, 10);
  
    // Find sum after the value is updated
    cout << "Updated sum of values in given range = "
         << query(n, 1, 3)
         << endl;
  
    return 0;
}

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Output:

Sum of values in given range = 15
Updated sum of values in given range = 45

Time Complexity:

  • For Tree Construction: O(N), There are (2n-1) nodes in the tree and value of every node is calculated once.
  • For Query: O(log N), To query a sum we processed atmost four nodes at every level and number of level is log N.
  • For Update: O(log N), To update the tree with lazy propagation is O(Log N) as we update the root of the tree and then update only that part of the tree whose ranges overlaps at each level.

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