Range Sum Query – Mutable LeetCode Solution

Problem – Range Sum Query – Mutable LeetCode Solution

Given an integer array nums, handle multiple queries of the following types:

  1. Update the value of an element in nums.
  2. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

  • NumArray(int[] nums) Initializes the object with the integer array nums.
  • void update(int index, int val) Updates the value of nums[index] to be val.
  • int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

Example 1:

["NumArray", "sumRange", "update", "sumRange"]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
[null, 9, null, 8]

Explanation NumArray numArray = new NumArray([1, 3, 5]); numArray.sumRange(0, 2); // return 1 + 3 + 5 = 9 numArray.update(1, 2); // nums = [1, 2, 5] numArray.sumRange(0, 2); // return 1 + 2 + 5 = 8


  • 1 <= nums.length <= 3 * 104
  • -100 <= nums[i] <= 100
  • 0 <= index < nums.length
  • -100 <= val <= 100
  • 0 <= left <= right < nums.length
  • At most 3 * 104 calls will be made to update and sumRange.

Range Sum Query – Mutable LeetCode Solution in Java

public class NumArray {

    class SegmentTreeNode {
        int start, end;
        SegmentTreeNode left, right;
        int sum;

        public SegmentTreeNode(int start, int end) {
            this.start = start;
            this.end = end;
            this.left = null;
            this.right = null;
            this.sum = 0;
    SegmentTreeNode root = null;
    public NumArray(int[] nums) {
        root = buildTree(nums, 0, nums.length-1);

    private SegmentTreeNode buildTree(int[] nums, int start, int end) {
        if (start > end) {
            return null;
        } else {
            SegmentTreeNode ret = new SegmentTreeNode(start, end);
            if (start == end) {
                ret.sum = nums[start];
            } else {
                int mid = start  + (end - start) / 2;             
                ret.left = buildTree(nums, start, mid);
                ret.right = buildTree(nums, mid + 1, end);
                ret.sum = ret.left.sum + ret.right.sum;
            return ret;
    void update(int i, int val) {
        update(root, i, val);
    void update(SegmentTreeNode root, int pos, int val) {
        if (root.start == root.end) {
           root.sum = val;
        } else {
            int mid = root.start + (root.end - root.start) / 2;
            if (pos <= mid) {
                 update(root.left, pos, val);
            } else {
                 update(root.right, pos, val);
            root.sum = root.left.sum + root.right.sum;

    public int sumRange(int i, int j) {
        return sumRange(root, i, j);
    public int sumRange(SegmentTreeNode root, int start, int end) {
        if (root.end == end && root.start == start) {
            return root.sum;
        } else {
            int mid = root.start + (root.end - root.start) / 2;
            if (end <= mid) {
                return sumRange(root.left, start, end);
            } else if (start >= mid+1) {
                return sumRange(root.right, start, end);
            }  else {    
                return sumRange(root.right, mid+1, end) + sumRange(root.left, start, mid);

Range Sum Query – Mutable LeetCode Solution in Python

    The idea here is to build a segment tree. Each node stores the left and right
    endpoint of an interval and the sum of that interval. All of the leaves will store
    elements of the array and each internal node will store sum of leaves under it.
    Creating the tree takes O(n) time. Query and updates are both O(log n).

#Segment tree node
class Node(object):
    def __init__(self, start, end):
        self.start = start
        self.end = end
        self.total = 0
        self.left = None
        self.right = None

class NumArray(object):
    def __init__(self, nums):
        initialize your data structure here.
        :type nums: List[int]
        #helper function to create the tree from input array
        def createTree(nums, l, r):
            #base case
            if l > r:
                return None
            #leaf node
            if l == r:
                n = Node(l, r)
                n.total = nums[l]
                return n
            mid = (l + r) // 2
            root = Node(l, r)
            #recursively build the Segment tree
            root.left = createTree(nums, l, mid)
            root.right = createTree(nums, mid+1, r)
            #Total stores the sum of all leaves under root
            #i.e. those elements lying between (start, end)
            root.total = root.left.total + root.right.total
            return root
        self.root = createTree(nums, 0, len(nums)-1)
    def update(self, i, val):
        :type i: int
        :type val: int
        :rtype: int
        #Helper function to update a value
        def updateVal(root, i, val):
            #Base case. The actual value will be updated in a leaf.
            #The total is then propogated upwards
            if root.start == root.end:
                root.total = val
                return val
            mid = (root.start + root.end) // 2
            #If the index is less than the mid, that leaf must be in the left subtree
            if i <= mid:
                updateVal(root.left, i, val)
            #Otherwise, the right subtree
                updateVal(root.right, i, val)
            #Propogate the changes after recursive call returns
            root.total = root.left.total + root.right.total
            return root.total
        return updateVal(self.root, i, val)

    def sumRange(self, i, j):
        sum of elements nums[i..j], inclusive.
        :type i: int
        :type j: int
        :rtype: int
        #Helper function to calculate range sum
        def rangeSum(root, i, j):
            #If the range exactly matches the root, we already have the sum
            if root.start == i and root.end == j:
                return root.total
            mid = (root.start + root.end) // 2
            #If end of the range is less than the mid, the entire interval lies
            #in the left subtree
            if j <= mid:
                return rangeSum(root.left, i, j)
            #If start of the interval is greater than mid, the entire inteval lies
            #in the right subtree
            elif i >= mid + 1:
                return rangeSum(root.right, i, j)
            #Otherwise, the interval is split. So we calculate the sum recursively,
            #by splitting the interval
                return rangeSum(root.left, i, mid) + rangeSum(root.right, mid+1, j)
        return rangeSum(self.root, i, j)

# Your NumArray object will be instantiated and called as such:
# numArray = NumArray(nums)
# numArray.sumRange(0, 1)
# numArray.update(1, 10)
# numArray.sumRange(1, 2)

Range Sum Query – Mutable LeetCode Solution in C++

class NumArray {
    // Check the constructor for the initialization of these variables.
    vector<int> seg; // Segment Tree to be stored in a vector.
    int n; // Length of the segment tree vector. 

    // Function to build the Segment Tree
    // This function will fill up the child values first
    // (left == right) will satisfy for the leaf values and they will be updated in segment tree
    // seg[pos]=seg[2*pos+1]+ seg[2*pos+2]; -> This will help populate all other intermediate nodes
    // as well as the root node with the "sum" of their child nodes.
    // Finally we have a segment tree which has all 'nodes' with sum values of their child.
    void buildTree(vector<int>& nums, int pos, int left, int right){
        if (left == right){
            seg[pos] = nums[left]; 
        int mid = (left+right)/2;
        buildTree(nums, 2*pos+1, left, mid);
        buildTree(nums, 2*pos+2, mid+1, right);
        seg[pos]=seg[2*pos+1]+ seg[2*pos+2];

    // Function to update a node in the segment tree
    // When a node is updated, then the change in the node value has to be propagated to the root
    // left, right -> represents the range of the node of segment tree. (Ex: [0, n-1] -> root)
    // pos       -> represents "position" in the segment tree data structure (Ex: 0 -> root)
    // Using left, right and pos -> we have all the information on the segment tree
    // Node at 'pos' in segment tree will have children at 2pos+1(left) and 2pos+2(right)

    // If index is less than left or more than right, then it is out of bound 
    //      for this node's range so we ignore it and return (This makes the algo O(logn))
    // If left==right==index, then we found the index, 
    //      update the value of the segment tree node & return
    // Otherwise, we need to find the index and we do this by checking child nodes (2pos+1, 2pos+2)
    //      update the segment tree pos with the updated child values' sum.
    //      This would help propagate the updated value of the chid indexes to the parent (through recursion)
    void updateUtil(int pos, int left, int right, int index, int val) {
        // no overlap
        if(index <left || index >right) return;
        // total overlap

        // partial overlap
        int mid=(left+right)/2;
        updateUtil(2*pos+1,left,mid,index,val); // left child
        updateUtil(2*pos+2,mid+1,right,index,val); // right child

    // Function to get the sum from the range [qlow, qhigh]
    // low, high -> represents the range of the node of segment tree. (Ex: [0, n-1] -> root)
    // pos       -> represents "position" in the segment tree data structure (Ex: 0 -> root)
    // Using low, high and pos -> we have all the information on the segment tree
    // Node at 'pos' in segment tree will have children at 2pos+1(left) and 2pos+2(right)
    // While searching for the range, there will be three cases: (Ex: arr: [-1, 4, 2, 0])
    //  - Total Overlap:    Return the value. (Ex: qlow, qhigh: 0,3 and low, high: 1,2)
    //  - No Overlap:       Return 0. (Ex: qlow, qhigh: 0,1 and low, high: 2,3)
    //  - Partial Overlap:  Search for it in both the child nodes and their ranges.
    //                      (Ex: Searching for 1,2 and node range is 0,1)
    int rangeUtil(int qlow, int qhigh, int low, int high, int pos){
        if (qlow <= low && qhigh>= high){ // total overlap
            return seg[pos];
        if (qlow > high || qhigh < low) { // no overlap
            return 0;
        // partial overlap
        int mid = low+(high-low)/2;
        return (rangeUtil(qlow, qhigh, low, mid, 2*pos+1) + rangeUtil(qlow, qhigh, mid+1, high, 2*pos+2));
    // Constructor for initializing the variables.
    NumArray(vector<int>& nums) {
        if(nums.size() > 0){
            n = nums.size();
            seg.resize(4*n,0);  // Maximum size of a segment tree for an array of size n is 4n
            buildTree(nums, 0, 0, n-1); // Build the segment tree
            // Print Segment Tree
            // for(int i=0;i<4*n;i++)
            //     cout<<seg[i]<<" ";
            // cout<<endl;
    // Update the segment Tree recurively using updateUtil
    void update(int index, int val) {
        updateUtil(0,0,n-1, index, val);
    // Get the sum for a specific range for the segment Tree
    int sumRange(int left, int right) {
        if(n==0)return 0;
        return rangeUtil(left, right, 0, n-1, 0); 
        // query from left to right while segment tree is now at 'root' (pos=0) and range(0, n-1)

 * Your NumArray object will be instantiated and called as such:
 * NumArray* obj = new NumArray(nums);
 * obj->update(index,val);
 * int param_2 = obj->sumRange(left,right);
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