Maximum array distance Sum

最新推荐文章于 2020-12-04 03:29:35 发布

原创最新推荐文章于 2020-12-04 03:29:35 发布 · 270 阅读

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本文介绍了一种寻找数组中最大距离和的方法。该方法通过分析数组特性，利用两个遍历过程来确定最大值，其中一个从前往后遍历，另一个从后往前遍历。这种方法不依赖额外的大内存空间，并且适用于各种数组数据类型。

Question: Given an array, find the maximum array distance sum. Array distance sum is defined as below. For any 0 <= p <= q, the array distance is A[p] + A[q] + (q - p)

For example, Given array {0, 2, -1}

A[0][0] = 0, A[0][1] = 3, A[0][2] = 1; A[1][1] = 2 + 2 + (1 - 1) = 4; A[1][2] = 2 + -1 + (2 - 1) = 2. A[2][2] = -1 + -1 + 0 = -2. Thus, the maximum array distance sum is: 4.

If allows extra O(N) memory space, we can do it by allocating an array for (A[p] - p), and another for (A[q] + q). Traverse the two array from front-end and end-front to get the maximum value.

If extra space was required for O(1), we need to analyse the question a bit. The array distance equals to A[p] - p + (A[q] + q). To get the max, we need to get max(A[p] - p) + max(A[q] + q), We can get that the max(A[p] - p) is the peak value in the array. However, max(A[q] + q) might exist after the peak.

#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int getMaxDistanceSum(vector<int>& array) {
    int forwardMax = INT_MIN;
    int backwardMax = INT_MIN;
    for(int i = 0; i < array.size(); ++i) {
       forwardMax = max(forwardMax, array[i] - i);
       backwardMax = max(backwardMax, array[i] + i);
    }
    return forwardMax + backwardMax;
}