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Create maximum-possible-number-by-binary-concatenation.cpp
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// Time: O(n * logr * logn)
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// Space: O(nlogr)
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// sort
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class Solution {
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public:
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int maxGoodNumber(vector<int>& nums) {
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const auto& to_binary = [](int x) {
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string result;
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for (; x; x >>= 1) {
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result.push_back(x & 1 ? '1' : '0');
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}
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reverse(begin(result), end(result));
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return result;
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};
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vector<string> bins(size(nums));
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for (int i = 0; i < size(nums); ++i) {
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bins[i] = to_binary(nums[i]);
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}
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sort(begin(bins), end(bins), [](const auto& a, const auto& b) {
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return (a + b) > (b + a);
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});
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string result;
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for (const auto& x : bins) {
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result += x;
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}
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return stoi(result, nullptr, 2);
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}
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};
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// Time: O(n! * nlogr)
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// Space: O(nlogr)
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// brute force
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class Solution2 {
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public:
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int maxGoodNumber(vector<int>& nums) {
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const auto& to_binary = [](int x) {
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string result;
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for (; x; x >>= 1) {
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result.push_back(x & 1 ? '1' : '0');
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}
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reverse(begin(result), end(result));
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return result;
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};
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vector<int> idxs(size(nums));
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iota(begin(idxs), end(idxs), 0);
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int result = 0;
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do {
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string curr;
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for (const auto& i : idxs) {
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curr += to_binary(nums[i]);
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}
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result = max(result, stoi(curr, nullptr, 2));
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} while (next_permutation(begin(idxs), end(idxs)));
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return result;
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}
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};

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