Count and Toggle Queries on a Binary Array

// C# program to implement toggle and 
// count queries on a binary array.
using System;

public class GFG{

    static readonly int MAX = 100000; 

    // segment tree to store count 
    // of 1's within range 
    static int []tree = new int[MAX]; 

    // bool type tree to collect the updates 
    // for toggling the values of 1 and 0 in 
    // given range 
    static bool []lazy = new bool[MAX]; 

    // function for collecting updates of toggling 
    // node --> index of current node in segment tree 
    // st --> starting index of current node 
    // en --> ending index of current node 
    // us --> starting index of range update query 
    // ue --> ending index of range update query 
    static void toggle(int node, int st, 
                    int en, int us, int ue) 
    { 
        // If lazy value is non-zero for current 
        // node of segment tree, then there are 
        // some pending updates. So we need 
        // to make sure that the pending updates 
        // are done before making new updates. 
        // Because this value may be used by 
        // parent after recursive calls (See last 
        // line of this function) 
        if (lazy[node]) 
        { 

            // Make pending updates using value 
            // stored in lazy nodes 
            lazy[node] = false; 
            tree[node] = en - st + 1 - tree[node]; 

            // checking if it is not leaf node 
            // because if it is leaf node then 
            // we cannot go further 
            if (st < en) 
            { 
                // We can postpone updating children 
                // we don't need their new values now. 
                // Since we are not yet updating children 
                // of 'node', we need to set lazy flags 
                // for the children 
                lazy[node << 1] = !lazy[node << 1]; 
                lazy[1 + (node << 1)] = !lazy[1 + (node << 1)]; 
            } 
        } 

        // out of range 
        if (st > en || us > en || ue < st) 
        { 
            return; 
        } 

        // Current segment is fully in range 
        if (us <= st && en <= ue) 
        { 
            // Add the difference to current node 
            tree[node] = en - st + 1 - tree[node]; 

            // same logic for checking leaf node or not 
            if (st < en) 
            { 
                // This is where we store values in lazy nodes, 
                // rather than updating the segment tree itself 
                // Since we don't need these updated values now 
                // we postpone updates by storing values in lazy[] 
                lazy[node << 1] = !lazy[node << 1]; 
                lazy[1 + (node << 1)] = !lazy[1 + (node << 1)]; 
            } 
            return; 
        } 

        // If not completely in rang, 
        // but overlaps, recur for children, 
        int mid = (st + en) / 2; 
        toggle((node << 1), st, mid, us, ue); 
        toggle((node << 1) + 1, mid + 1, en, us, ue); 

        // And use the result of children 
        // calls to update this node 
        if (st < en) 
        { 
            tree[node] = tree[node << 1] + 
                        tree[(node << 1) + 1]; 
        } 
    } 

    /* node --> Index of current node in the segment tree. 
        Initially 0 is passed as root is always at' 
        index 0 
    st & en --> Starting and ending indexes of the 
                segment represented by current node, 
                i.e., tree[node] 
    qs & qe --> Starting and ending indexes of query 
                range */
    // function to count number of 1's 
    // within given range 
    static int countQuery(int node, int st, 
                        int en, int qs, int qe) 
    { 
        // current node is out of range 
        if (st > en || qs > en || qe < st) 
        { 
            return 0; 
        } 

        // If lazy flag is set for current 
        // node of segment tree, then there 
        // are some pending updates. So we 
        // need to make sure that the pending 
        // updates are done before processing 
        // the sub sum query 
        if (lazy[node]) 
        { 
            // Make pending updates to this node. 
            // Note that this node represents sum 
            // of elements in arr[st..en] and 
            // all these elements must be increased 
            // by lazy[node] 
            lazy[node] = false; 
            tree[node] = en - st + 1 - tree[node]; 

            // checking if it is not leaf node because if 
            // it is leaf node then we cannot go further 
            if (st < en) 
            { 
                // Since we are not yet updating children os si, 
                // we need to set lazy values for the children 
                lazy[node << 1] = !lazy[node << 1]; 
                lazy[(node << 1) + 1] = !lazy[(node << 1) + 1]; 
            } 
        } 

        // At this point we are sure that pending 
        // lazy updates are done for current node. 
        // So we can return value If this segment 
        // lies in range 
        if (qs <= st && en <= qe) 
        { 
            return tree[node]; 
        } 

        // If a part of this segment overlaps 
        // with the given range 
        int mid = (st + en) / 2; 
        return countQuery((node << 1), st, mid, qs, qe) + 
            countQuery((node << 1) + 1, mid + 1, en, qs, qe); 
    } 

    // Driver Code 
    public static void Main() 
    { 
        int n = 5; 
        toggle(1, 0, n - 1, 1, 2); // Toggle 1 2 
        toggle(1, 0, n - 1, 2, 4); // Toggle 2 4 

        Console.WriteLine(countQuery(1, 0, n - 1, 2, 3)); // Count 2 3 

        toggle(1, 0, n - 1, 2, 4); // Toggle 2 4 

        Console.WriteLine(countQuery(1, 0, n - 1, 1, 4)); // Count 1 4 
    } 
} 

/*This code is contributed by PrinciRaj1992*/