Array Inversion and Anti-Inversion

Below is a concise explanation and a corrected C++ solution that produces the required results:

Understanding the Problem

We have an array $A$ of size $N$. We want to remove exactly one element from $A$ (at some index $i$) so that in the resulting array $A'$ (of size $N-1$) the quantity

is maximized.

• An inversion pair $(p, q)$ in $A'$ is one where $p < q$ and $A'_p > A'_q.$
• An anti-inversion pair $(p, q)$ in $A'$ is one where $p < q$ and $A'_p < A'_q.$

Among all indices that yield the same maximum sum, we must choose the largest index.

Key Observation

For any array of size $M$, the total number of index-pairs $(p,q)$ with $p < q$ is

Such a pair is either:

1. An inversion ( $A'_p > A'_q$ ),
2. An anti-inversion ( $A'_p < A'_q$ ), or
3. Neither, which can only happen if $A'_p = A'_q$.

Hence,

After removing one element, our new array size is $N - 1$. Therefore,

Counting “equal pairs”

Let $\text{freq}[x]$ be the frequency of value $x$ in the original array. The number of pairs of equal elements in the original array is:

When you remove one element of value $v$, its frequency in the new array becomes $\text{freq}[v] - 1$. Consequently, the count of equal pairs for value $v$ changes from

The difference is

Hence, removing one element of value $v$ reduces the number of equal-pairs by $\text{freq}[v] - 1$.

Putting It All Together

1. Let $$\text{totalEqualPairs} = \sum_{x} \binom{\text{freq}[x]}{2}.$$
2. For the new array of size $N-1$, the total possible pairs is $$\binom{N-1}{2} = \frac{(N-1)(N-2)}{2}.$$
3. If we remove an element whose value is $v$, the new number of equal pairs becomes $$\text{totalEqualPairs} \;-\; \bigl(\text{freq}[v] - 1\bigr).$$
4. Therefore, the sum of inversions + anti-inversions in the new array is $$\binom{N-1}{2} \;-\; \Bigl( \text{totalEqualPairs} - (\text{freq}[v] - 1) \Bigr) = \underbrace{\binom{N-1}{2} - \text{totalEqualPairs}}_{\text{constant } K} \;+\; \bigl(\text{freq}[v] - 1\bigr).$$ Let $$K = \binom{N-1}{2} - \text{totalEqualPairs}.$$ Then the result for removing a value $v$ is $$\text{Result}(v) = K \;+\; \bigl(\text{freq}[v] - 1\bigr).$$

To maximize $\text{Result}(v)$, we want to remove an element whose value $v$ has the largest frequency $\text{freq}[v]$. If multiple elements have the same frequency, we pick the largest index among them, as the problem demands.

Correct C++ Implementation

cpp
#include <bits/stdc++.h>
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;
    vector<long long> A(N);
    for(int i = 0; i < N; i++){
        cin >> A[i];
    }

    // 1) Compute frequencies
    unordered_map<long long, long long> freq;
    freq.reserve(N);  // small optimization
    freq.max_load_factor(0.7f);
    for(auto &val : A){
        freq[val]++;
    }

    // 2) Compute total number of equal-pairs in the original array
    long long totalEqualPairs = 0;
    for(auto &p : freq){
        long long c = p.second;
        totalEqualPairs += (c * (c - 1)) / 2;
    }

    // 3) Precompute K = C(N-1, 2) - totalEqualPairs
    long long totalPairs = (long long)(N-1) * (N-2) / 2; 
    long long K = totalPairs - totalEqualPairs;

    // We want to find index i that maximizes sum = K + (freq[A[i]] - 1).
    // Tie-break: choose largest i if sums are same.
    long long maxSum = LLONG_MIN;
    int answerIndex = -1;

    // Iterate from left to right, but *update* if we find >= (so we end with largest index).
    // Alternatively, iterate from right to left and pick the first maximum we see.
    // Either way works, but let's do left->right with ">=":

    for(int i = 0; i < N; i++){
        long long val = A[i];
        long long currentSum = K + (freq[val] - 1);
        if(currentSum > maxSum){
            maxSum = currentSum;
            answerIndex = i + 1; // +1 because problem statements often use 1-based index
        }
        else if(currentSum == maxSum){
            // tie-break: pick the larger index
            // i+1 will be >= answerIndex always if i is to the right
            answerIndex = i + 1;
        }
    }

    cout << answerIndex << "\n";
    return 0;
}

Explanation of the Steps

1. Frequencies: Count how often each integer appears.
2. Total Equal Pairs: For each distinct value $x$ with frequency $\text{freq}[x]$, it contributes $\text{freq}[x](\text{freq}[x] - 1)/2$ “equal” pairs.
3. Compute $K$: $$K = \binom{N-1}{2} - \sum_x \binom{\text{freq}[x]}{2}.$$ This piece is constant for the whole array.
4. Iterate Over Each Index:
   • Let $v = A[i]$. Removing $A[i]$ changes the frequency of $v$ from $\text{freq}[v]$ to $\text{freq}[v] - 1$.
   • The resulting sum becomes $K + (\text{freq}[v] - 1)$.
   • We keep track of the maximum such sum, and if multiple indices yield that maximum, choose the largest index.

Checking the Provided Examples

1. Example 1
   Input: $N=4,\, A=[1,\,2,\,2,\,1]$
   Output should be 4.

   • Whichever element we remove, we get the same final sum (2). Hence the largest index is 4.

2. Example 2
   Input: $N=6,\, A=[1,\,2,\,2,\,3,\,4,\,6]$
   Output should be 3.

   • Removing one of the two “2”s yields the highest final sum, and among the two positions where $A[i] = 2$ (indices 2 and 3 in 1-based), 3 is larger.

This code will pass these and other test cases correctly, fixing the typical off-by-one and logic bugs that often appear in manual or naive solutions.