본인의 풀이는 현재 보드에서 나올 수 있는 모든 가지수를 구해서 최소뒤집는 횟수를 리턴해주는 방식이다.
하지만 좋은 풀이가 아니므로 아래의 다른 코드를 참고하자.

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

int answer = -1;
int n,m;

void flipRow(int idx, vector<vector<int>> *board) {
    for(int i=0; i<m; i++) {
        if ((*board)[idx][i] == 0) (*board)[idx][i] = 1;
        else (*board)[idx][i] = 0;
    }
}

void flipCol(int idx, vector<vector<int>> *board) {
    for(int i=0; i<n; i++) {
        if ((*board)[i][idx] == 0) (*board)[i][idx] = 1;
        else (*board)[i][idx] = 0;
    }
}

void isSameBoard(vector<vector<int>> board, vector<int> flipRowIdx, vector<int> flipColIdx, vector<vector<int>> target) {
    for (auto idx : flipRowIdx) flipRow(idx, &board);
    for (auto idx : flipColIdx) flipCol(idx, &board);
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            if (board[i][j] != target[i][j]) return ;
    int res = flipRowIdx.size() + flipColIdx.size();
    if (answer == -1) answer = res;
    else answer = min(answer, res);
}

void dfs(int idx, int num, vector<int> tmp, vector<vector<int>> *flipIdxs) {
    (*flipIdxs).push_back(tmp);
    for (int i=idx; i<num; i++) {
        tmp.push_back(i);
        dfs(i+1, num, tmp, flipIdxs);
        tmp.pop_back();
    }
}

int solution(vector<vector<int>> beginning, vector<vector<int>> target) {
    n = beginning.size();
    m = beginning[0].size();
    vector<vector<int>> flipRowIdxs, flipColIdxs;
    dfs(0,n,{}, &flipRowIdxs);
    dfs(0,m,{}, &flipColIdxs);
    for (auto flipRowIdx : flipRowIdxs) {
        for (auto flipColIdx : flipColIdxs) {
            isSameBoard(beginning, flipRowIdx, flipColIdx, target);
        }
    }
    return answer;
}

좋은 코드

#include <vector>
#include <iostream>

using namespace std;

int solution(vector<vector<int>> beginning, vector<vector<int>> target) {
    int answer=0;
    int r_size=beginning.size(), c_size=beginning[0].size();
    vector<vector<int>> board(r_size,vector<int>(c_size));
    for(int i=0; i<r_size; i++)
        for(int j=0; j<c_size; j++)
            board[i][j]=beginning[i][j]^target[i][j];
    int cnt=0, sum=0;
    for(int i=0; i<c_size; i++) sum+=board[0][i];
    for(int i=1; i<r_size; i++) {
        bool o_flag=true, r_flag=true;
        for(int j=0; j<c_size; j++) {
            if((board[i][j]^0)!=board[0][j]) o_flag=false;
            if((board[i][j]^1)!=board[0][j]) r_flag=false;
        }
        if (!o_flag) {
            if(!r_flag) return -1;
            else cnt++;
        }
    }
    answer=min(cnt+sum,r_size-cnt+c_size-sum);
    return answer;
}