通过翻转C ++中的子数组来最大化0
问题陈述
给定二进制数组，请在允许翻转子数组的情况下找到数组中的最大零个数。翻转操作将所有0切换为1s，将1s切换为0s
如果arr1={1、1、0、0、0、0、0}
如果将前2个1翻转为0，则可以得到大小为7的子数组，如下所示：
{0, 0, 0, 0, 0, 0, 0}算法
1. Consider all subarrays and find a subarray with maximum value of (count of 1s) – (count of 0s) 2. Considers this value be maxDiff. Finally return count of zeros in original array + maxDiff.
示例
#include <bits/stdc++.h>
using namespace std;
int getMaxSubArray(int *arr, int n){
int maxDiff = 0;
int zeroCnt = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] == 0) {
++zeroCnt;
}
int cnt0 = 0;
int cnt1 = 0;
for (int j = i; j < n; ++j) {
if (arr[j] == 1) {
++cnt1;
}
else {
++cnt0;
}
maxDiff = max(maxDiff, cnt1 - cnt0);
}
}
return zeroCnt + maxDiff;
}
int main(){
int arr[] = {1, 1, 0, 0, 0, 0, 0};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum subarray size = " << getMaxSubArray(arr, n) << endl;
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
Maximum subarray size = 7