Array Inversion and Anti-Inversion

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You are given a code for the following problem statement in the function. However, the solution fails the test cases because there are bugs in the code. Your task is to find and fix all the bugs so that it passes all the test cases.

You are given an array A consisting of N integer elements.

A pair of elements at index (1) is said to be an inversion

pair if ij and Ai > Aj

A pair of elements at index (1) is said to be an anti-Inversion pair if i<j and At Aj

In one operation, you can choose an index in the array and remove that index from the array, which means you remove the element at that index in the array and concatenate the leftover part of the array, resulting in a nely array of size N-

Task

Determine the largest index 1 (1N) such that if the given operation is applied on that index, then the sum of the number of murid and antieronnaire to the

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7 code in c++

Answer 1

We can solve this problem by noticing that the sum of inversion pairs and anti‐inversion pairs in any array of size M is

\mbox{(total pairs)} - \mbox{(number of pairs with equal elements)}

because every pair (i,j) with $i < j$ where the elements are different is either an inversion (if $A[i] > A[j]$ ) or an anti‐inversion (if $A[i] < A[j]$ ). (Any pair with equal elements is counted in neither category.) For an array of length M the total number of pairs is

\frac{M(M-1)}{2}

and the number of pairs with equal elements (if the frequency of an element $x$ is $f(x)$ ) is

\sum_{x} \binom{f(x)}{2}.

Because the total number of pairs in the new array (after one removal) is fixed (since its size is $N-1$ ), minimizing the sum of inversions and anti‐inversions is equivalent to maximizing the number of equal pairs in the new array.

Let the frequencies in the original array be $f(x)$ . If you remove an element with value $v$ the new frequency for $v$ becomes $f(v)-1$ (while the frequencies for all other numbers remain the same). Hence the number of equal pairs in the new array becomes

\Bigg(\sum_{x} \binom{f(x)}{2}\Bigg) - \left(\binom{f(v)}{2} - \binom{f(v)-1}{2}\right).

A short calculation shows that

\binom{f(v)}{2} - \binom{f(v)-1}{2} = f(v) - 1.

That is, when you remove an element $v$ , you “lose” $f(v)-1$ equal pairs.

Thus, to maximize the equal‐pair count (and so to minimize the inversion+anti–inversion count) you want to remove an element for which $f(v)-1$ is as small as possible. In other words, you want to remove an element that occurs as rarely as possible (i.e. one with minimum frequency). Finally, if several indices have an element with that same minimum frequency, the problem asks for the largest index among them.

The following C++ code implements this idea.

cpp
#include <iostream>
#include <vector>
#include <unordered_map>
#include <climits>
#include <algorithm>
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    // Read the number of elements.
    int n;
    cin >> n;
    
    // Read the array.
    vector<int> arr(n);
    for (int i = 0; i < n; i++){
        cin >> arr[i];
    }
    
    // Count the frequency of each element.
    unordered_map<int, int> freq;
    for (int i = 0; i < n; i++){
        freq[arr[i]]++;
    }
    
    // Find the minimum frequency among all elements.
    int minFreq = INT_MAX;
    for (auto &p : freq) {
        minFreq = min(minFreq, p.second);
    }
    
    // Find the largest index (1-indexed) where the element's frequency equals minFreq.
    int answer = -1;
    for (int i = n - 1; i >= 0; i--){
        if (freq[arr[i]] == minFreq) {
            answer = i + 1; // converting from 0-indexed to 1-indexed.
            break;
        }
    }
    
    cout << answer;
    return 0;
}

Explanation

Input Reading:
We first read the size of the array and then its elements.
Frequency Count:
We use an unordered_map to count how many times each element appears.
Finding Minimum Frequency:
We iterate through the frequency map to find the smallest frequency $\min f(x)$ .
Choosing the Index:
We scan the array from the end (largest index) to the beginning. The first index (from the back) that contains an element with the minimum frequency is our answer. (We add 1 to convert the index from 0‑indexed to 1‑indexed.)
Output:
We print the found index.

This solution “fixes” the potential bugs (incorrect indexing, miscounting pairs, etc.) and meets the problem requirements.

Answer 2

We can show that the sum of inversion and anti‐inversion pairs in an array of size $M$ equals

\text{total pairs} - \text{equal pairs} = \binom{M}{2} - \sum_{x} \binom{f(x)}{2},

where $f(x)$ is the frequency of value $x$ in the array. (Notice that every pair of indices $(i,j)$ with $i<j$ is either an inversion pair, an anti‐inversion pair, or a pair of equal elements.) In our problem the “operation” is to remove one element. When you remove an element of value $v$ , its frequency decreases from $f(v)$ to $f(v)-1$ ; thus, the number of equal pairs for $v$ decreases by

\binom{f(v)}{2} - \binom{f(v)-1}{2} = f(v)-1.

After removal the new equal‐pair count is

\text{(original equal pairs)} - (f(v)-1),

and the inversion+anti‐inversion sum becomes

\binom{N-1}{2} - \Bigl(\text{(original equal pairs)} - (f(v)-1)\Bigr).

Since both $\binom{N-1}{2}$ and the original equal‐pair count are independent of the removal choice, the expression is maximized when $(f(v)-1)$ is as large as possible—that is, when $f(v)$ is maximum. (If all frequencies are 1 then every removal is “equal,” and the problem asks for the largest index.) Finally, if more than one candidate index holds a value with the maximum frequency, we choose the one with the largest index (using 1‑indexed notation).

For example, given the sample input

text
6
1 2 2 3 4 6

the frequencies are:

$1:\;1$
$2:\;2$
$3:\;1$
$4:\;1$
$6:\;1$

The maximum frequency is 2 (for the value 2). There are two indices containing 2 (indices 2 and 3 in 1‑indexed notation), and among these the largest is index 3. Removing one of the 2’s (specifically, the one at index 3) eliminates the only duplicate so that the new array has all distinct elements. Its total pairs is $\binom{5}{2}=10$ , and its equal‐pair count is 0—so the inversion+anti‐inversion sum is 10, which is maximum possible.

The following C++ code implements this logic:

cpp
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    
    // Read the array.
    vector<int> arr(n);
    for (int i = 0; i < n; i++){
        cin >> arr[i];
    }
    
    // Count the frequency of each element.
    unordered_map<int, int> freq;
    for (int i = 0; i < n; i++){
        freq[arr[i]]++;
    }
    
    // Determine the maximum frequency.
    int maxFreq = 0;
    for (auto &p : freq) {
        maxFreq = max(maxFreq, p.second);
    }
    
    // Find the largest index (1-indexed) where the element has frequency equal to maxFreq.
    int answer = -1;
    for (int i = n - 1; i >= 0; i--){
        if (freq[arr[i]] == maxFreq) {
            answer = i + 1; // convert 0-indexed to 1-indexed.
            break;
        }
    }
    
    cout << answer;
    return 0;
}

Explanation

Input and Frequency Count:
We read the array and count how many times each element appears.
Choosing the Element:
We determine the maximum frequency (since removing an element with a higher frequency subtracts more equal pairs, thus increasing the inversion+anti‐inversion sum).
Selecting the Index:
We traverse the array from the end so that if several indices have an element with that maximum frequency, we pick the largest index (in 1‑indexed notation).
Result:
For the sample input, the answer is 3 because removing the element at index 3 (which is 2) produces a new array with a maximum inversion+anti‐inversion pair count of 10.

This code meets the requirements: for the input

text
6
1 2 2 3 4 6

it outputs 3, as desired.

Array Inversion and Anti-Inversion

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