Translate the following into Chinese: C++ Query to answer the number of 1's and 0's to the left of a given index

Discuss a question to answer a query on a given array. For example, for each query index, we need to find the number of 1's and 0's to the left of the index.

Input: arr[ ] = { 0, 1, 1, 1, 0, 0, 0, 1, 0, 0}, queries[ ] = { 2, 4, 1, 0, 5 }
Output:
query 1: zeros = 1,ones = 1
query 2: zeros = 1,ones = 3
query 3: zeros = 1,ones = 0
query 4: zeros = 0,ones = 0
query 5: zeros = 2,ones = 3

Input: arr[ ] = { 0, 0, 1, 1, 1, 0, 1, 0, 0, 1 }, queries[ ] = { 3, 2, 6 }
Output:
query 1: zeros = 2,ones = 1
query 2: zeros = 2,ones = 0
query 3: zeros = 3,ones = 3

Ways to find a solution

Naive way

A simple way to solve this problem is to iterate through the array to the index of the query and check each element; if it is 0, Then increment the zero counter by 1, otherwise increment the zero counter by 1.

Example

#include <bits/stdc++.h>
using namespace std;
int main(){
    int nums[] = {1, 0, 0, 1, 1, 0, 0, 1, 0, 0};
    int queries[] =  { 2, 4, 1, 0, 5 };
    int qsize = sizeof(queries) / sizeof(queries[0]);
    int zeros=0,ones=0;
    // loop for running each query.
    for(int i = 0;i<qsize;i++){
        //counting zeros and ones
        for(int j = 0;j<queries[i];j++){
            if(nums[j]==0)
            zeros++;
            else
            ones++;
        }
        cout << "\nquery " << i+1 << ": zeros = " << zeros << ",ones = " << ones;
        zeros=0;
        ones=0;
    }
    return 0;
}

Output

query 1: zeros = 1,ones = 1
query 2: zeros = 2,ones = 2
query 3: zeros = 0,ones = 1
query 4: zeros = 0,ones = 0
query 5: zeros = 2,ones = 3

Efficient method

In the previous method, every time we start from the 0th index to calculate the new query 1 and 0.

Another way is to count 0 and 1 first. appears to the left of each index, stores them in an array, and returns the answer based on the index written in the query.

Example

#include <bits/stdc++.h>
using namespace std;
int main(){
    int nums[] = {1, 0, 0, 1, 1, 0, 0, 1, 0, 0};
    int queries[] =  { 2, 4, 1, 0, 5 };
    int n = sizeof(nums) / sizeof(nums[0]);
    int arr[n][2];
    int zeros = 0, ones = 0;
    // traverse through the nums array.
    for (int i = 0; i < n; i++) {
        // store the number of zeros and ones in arr.
        arr[i][0] = zeros;
        arr[i][1] = ones;
        // increment variable according to condition
        if (nums[i]==0)
            zeros++;
        else
            ones++;
    }
    int qsize = sizeof(queries) / sizeof(queries[0]);
        for (int i = 0; i < qsize; i++)
        cout << "\nquery " << i+1 << ": zeros = " << arr[queries[i]][0] << ",ones ="    << arr[queries[i]][1];
    return 0;
}

Output

query 1: zeros = 1,ones =1
query 2: zeros = 2,ones =2
query 3: zeros = 0,ones =1
query 4: zeros = 0,ones =0
query 5: zeros = 2,ones =3

Conclusion

In this tutorial we discussed about returning the index left for every query in a given array The number of 1's and 0's. We discussed simple and effective ways to solve this problem. We also discussed a C program to solve this problem and we can implement it using programming languages ​​like C, Java, Python etc. We hope you found this tutorial helpful.

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