Problem p02599 (C++) — buggy submission, verdict: Time Limit Exceeded

// C++ code to find number of distinct numbers
// in a subarray
#include <bits/stdc++.h>
using namespace std;

const int MAX = 1000001;

// structure to store queries
struct Query {
  int l, r, idx;
};

// cmp function to sort queries according to r
bool cmp(Query x, Query y) { return x.r < y.r; }

// updating the bit array
void update(int idx, int val, int bit[], int n) {
  for (; idx <= n; idx += idx & -idx)
    bit[idx] += val;
}

// querying the bit array
int query(int idx, int bit[], int n) {
  int sum = 0;
  for (; idx > 0; idx -= idx & -idx)
    sum += bit[idx];
  return sum;
}

void answeringQueries(int arr[], int n, Query queries[], int q) {
  // initialising bit array
  int bit[n + 1];
  memset(bit, 0, sizeof(bit));

  // holds the rightmost index of any number
  // as numbers of a[i] are less than or equal to 10^6
  int last_visit[MAX];
  memset(last_visit, -1, sizeof(last_visit));

  // answer for each query
  int ans[q];
  int query_counter = 0;
  for (int i = 0; i < n; i++) {
    // If last visit is not -1 update -1 at the
    // idx equal to last_visit[arr[i]]
    if (last_visit[arr[i]] != -1)
      update(last_visit[arr[i]] + 1, -1, bit, n);

    // Setting last_visit[arr[i]] as i and updating
    // the bit array accordingly
    last_visit[arr[i]] = i;
    update(i + 1, 1, bit, n);

    // If i is equal to r of any query store answer
    // for that query in ans[]
    while (query_counter < q && queries[query_counter].r == i) {
      ans[queries[query_counter].idx] =
          query(queries[query_counter].r + 1, bit, n) -
          query(queries[query_counter].l, bit, n);
      query_counter++;
    }
  }

  // print answer for each query
  for (int i = 0; i < q; i++)
    cout << ans[i] << endl;
}

// driver code
int main() {
  int n, Q;
  cin >> n >> Q;
  int a[n];
  Query queries[Q];
  for (int i = 0; i < n; i++)
    cin >> a[i];
  for (int i = 0; i < Q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    queries[i] = {l, r, i};
  }
  sort(queries, queries + Q, cmp);
  answeringQueries(a, n, queries, Q);
  return 0;
}


Error: TLE

Fixed submission:

// Program to compute no. of different elements
// of ranges for different range queries
#include <bits/stdc++.h>
using namespace std;

// Used in frequency array (maximum value of an
// array element).
const int MAX = 500001;

// Variable to represent block size. This is made
// global so compare() of sort can use it.
int block;

// Structure to represent a query range and to store
// index and result of a particular query range
struct Query {
  int L, R, index, result;
};

// Function used to sort all queries so that all queries
// of same block are arranged together and within a block,
// queries are sorted in increasing order of R values.
bool compare(Query x, Query y) {
  // Different blocks, sort by block.
  if (x.L / block != y.L / block)
    return x.L / block < y.L / block;

  // Same block, sort by R value
  return x.R < y.R;
}

// Function used to sort all queries in order of their
// index value so that results of queries can be printed
// in same order as of input
bool compare1(Query x, Query y) { return x.index < y.index; }

// calculate distinct elements of all query ranges.
// m is number of queries n is size of array a[].
void queryResults(int a[], int n, Query q[], int m) {
  // Find block size
  block = (int)sqrt(n);

  // Sort all queries so that queries of same
  // blocks are arranged together.
  sort(q, q + m, compare);

  // Initialize current L, current R and current
  // different elements
  int currL = 0, currR = 0;
  int curr_Diff_elements = 0;

  // Initialize frequency array with 0
  int freq[MAX] = {0};

  // Traverse through all queries
  for (int i = 0; i < m; i++) {

    // L and R values of current range
    int L = q[i].L, R = q[i].R;

    // Remove extra elements of previous range.
    // For example if previous range is [0, 3]
    // and current range is [2, 5], then a[0]
    // and a[1] are subtracted
    while (currL < L) {

      // element a[currL] is removed
      freq[a[currL]]--;
      if (freq[a[currL]] == 0)
        curr_Diff_elements--;

      currL++;
    }

    // Add Elements of current Range
    // Note:- during addition of the left
    // side elements we have to add currL-1
    // because currL is already in range
    while (currL > L) {
      freq[a[currL - 1]]++;

      // include a element if it occurs first time
      if (freq[a[currL - 1]] == 1)
        curr_Diff_elements++;

      currL--;
    }
    while (currR <= R) {
      freq[a[currR]]++;

      // include a element if it occurs first time
      if (freq[a[currR]] == 1)
        curr_Diff_elements++;

      currR++;
    }

    // Remove elements of previous range. For example
    // when previous range is [0, 10] and current range
    // is [3, 8], then a[9] and a[10] are subtracted
    // Note:- Basically for a previous query L to R
    // currL is L and currR is R+1. So during removal
    // of currR remove currR-1 because currR was
    // never included
    while (currR > R + 1) {

      // element a[currL] is removed
      freq[a[currR - 1]]--;

      // if occurrence of a number is reduced
      // to zero remove it from list of
      // different elements
      if (freq[a[currR - 1]] == 0)
        curr_Diff_elements--;

      currR--;
    }
    q[i].result = curr_Diff_elements;
  }
}

// print the result of all range queries in
// initial order of queries
void printResults(Query q[], int m) {
  sort(q, q + m, compare1);
  for (int i = 0; i < m; i++) {
    cout << q[i].result << endl;
  }
}

// Driver program
int main() {
  int n, Q;
  cin >> n >> Q;
  int a[n];
  Query q[Q];
  for (int i = 0; i < n; i++)
    cin >> a[i];
  for (int i = 0; i < Q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    q[i] = {l, r, i, 0};
  }
  queryResults(a, n, q, Q);
  printResults(q, Q);
  return 0;
}