mirror of https://github.com/doocs/leetcode.git
feat: add solutions to lc problem: No.3007 (#3437)
No.3007.Maximum Number That Sum of the Prices Is Less Than or Equal to K
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Libin YANG
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@ -245,7 +245,17 @@ tags:
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<!-- solution:start -->
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### 方法一
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### 方法一：二分查找 + 数位 DP
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我们注意到，如果 $\textit{num}$ 增大，数字 $1$ 到 $\textit{num}$ 的总价值也会增大。因此，我们可以使用二分查找的方法找到最大的廉价数字。
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我们定义二分查找的左边界 $l = 1$，由于每 $2^x + 1$ 个数中至少有一个数字是有价值的，而总价值不超过 $10^15$，因此我们可以设定二分查找的右边界 $r = 10^{18}$。
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接下来，我们进行二分查找，对于每一个 $\textit{mid}$，我们使用数位 DP 的方法计算出 $1$ 到 $\textit{mid}$ 的总价值，如果总价值不超过 $k$，则说明 $\textit{mid}$ 是一个廉价数字，我们将左边界 $l$ 更新为 $\textit{mid}$，否则我们将右边界 $r$ 更新为 $\textit{mid} - 1$。
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最后，我们返回左边界 $l$ 即可。
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时间复杂度 $O(\log^2 k)$，空间复杂度 $O(\log k)$。
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<!-- tabs:start -->
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@ -332,7 +342,7 @@ public:
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ll l = 1, r = 1e17;
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ll num = 0;
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ll f[65][65];
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function<ll(int, int, bool)> dfs = [&](int pos, int cnt, bool limit) -> ll {
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auto dfs = [&](auto&& dfs, int pos, int cnt, bool limit) -> ll {
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if (pos == 0) {
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return cnt;
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}
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@ -342,7 +352,7 @@ public:
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int up = limit ? num >> (pos - 1) & 1 : 1;
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ll ans = 0;
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for (int i = 0; i <= up; ++i) {
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ans += dfs(pos - 1, cnt + (i == 1 && pos % x == 0), limit && i == up);
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ans += dfs(dfs, pos - 1, cnt + (i == 1 && pos % x == 0), limit && i == up);
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}
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if (!limit) {
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f[pos][cnt] = ans;
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@ -354,7 +364,7 @@ public:
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num = mid;
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memset(f, -1, sizeof(f));
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int pos = 64 - __builtin_clzll(mid);
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if (dfs(pos, 0, true) <= k) {
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if (dfs(dfs, pos, 0, true) <= k) {
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l = mid;
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} else {
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r = mid - 1;
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@ -416,6 +426,56 @@ func findMaximumNumber(k int64, x int) int64 {
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}
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```
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#### TypeScript
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```ts
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function findMaximumNumber(k: number, x: number): number {
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let [l, r] = [1n, 10n ** 17n];
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let num: bigint;
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const f: bigint[][] = Array.from({ length: 65 }, () => Array(65).fill(-1n));
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const dfs = (pos: number, cnt: number, limit: boolean): bigint => {
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if (pos === 0) {
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return BigInt(cnt);
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}
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if (!limit && f[pos][cnt] !== -1n) {
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return f[pos][cnt];
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}
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let ans: bigint = 0n;
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let up: number = 1;
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if (limit) {
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up = Number((num >> BigInt(pos - 1)) & 1n);
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}
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for (let i = 0; i <= up; i++) {
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let v: number = cnt;
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if (i === 1 && pos % x === 0) {
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v++;
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}
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ans += dfs(pos - 1, v, limit && i === up);
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}
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if (!limit) {
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f[pos][cnt] = ans;
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}
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return ans;
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};
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while (l < r) {
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let mid: bigint = (l + r + 1n) >> 1n;
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num = mid;
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let m: number = num.toString(2).length;
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for (let i = 0; i < f.length; i++) {
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f[i].fill(-1n);
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}
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if (dfs(m, 0, true) <= BigInt(k)) {
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l = mid;
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} else {
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r = mid - 1n;
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}
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}
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return Number(l);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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@ -249,7 +249,17 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Binary Search + Digit DP
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We notice that if $\textit{num}$ increases, the total value from $1$ to $\textit{num}$ also increases. Therefore, we can use a binary search method to find the largest cheap number.
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We define the left boundary of the binary search as $l = 1$. Since there is at least one valuable number in every $2^x + 1$ numbers, and the total value does not exceed $10^{15}$, we can set the right boundary of the binary search as $r = 10^{18}$.
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Next, we perform a binary search. For each $\textit{mid}$, we use the digit DP method to calculate the total value from $1$ to $\textit{mid}$. If the total value does not exceed $k$, it means $\textit{mid}$ is a cheap number, and we update the left boundary $l$ to $\textit{mid}$. Otherwise, we update the right boundary $r$ to $\textit{mid} - 1$.
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Finally, we return the left boundary $l$.
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The time complexity is $O(\log^2 k)$, and the space complexity is $O(\log k)$.
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<!-- tabs:start -->
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@ -336,7 +346,7 @@ public:
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ll l = 1, r = 1e17;
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ll num = 0;
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ll f[65][65];
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function<ll(int, int, bool)> dfs = [&](int pos, int cnt, bool limit) -> ll {
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auto dfs = [&](auto&& dfs, int pos, int cnt, bool limit) -> ll {
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if (pos == 0) {
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return cnt;
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}
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@ -346,7 +356,7 @@ public:
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int up = limit ? num >> (pos - 1) & 1 : 1;
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ll ans = 0;
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for (int i = 0; i <= up; ++i) {
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ans += dfs(pos - 1, cnt + (i == 1 && pos % x == 0), limit && i == up);
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ans += dfs(dfs, pos - 1, cnt + (i == 1 && pos % x == 0), limit && i == up);
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}
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if (!limit) {
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f[pos][cnt] = ans;
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@ -358,7 +368,7 @@ public:
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num = mid;
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memset(f, -1, sizeof(f));
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int pos = 64 - __builtin_clzll(mid);
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if (dfs(pos, 0, true) <= k) {
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if (dfs(dfs, pos, 0, true) <= k) {
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l = mid;
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} else {
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r = mid - 1;
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@ -420,6 +430,56 @@ func findMaximumNumber(k int64, x int) int64 {
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}
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```
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#### TypeScript
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```ts
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function findMaximumNumber(k: number, x: number): number {
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let [l, r] = [1n, 10n ** 17n];
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let num: bigint;
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const f: bigint[][] = Array.from({ length: 65 }, () => Array(65).fill(-1n));
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const dfs = (pos: number, cnt: number, limit: boolean): bigint => {
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if (pos === 0) {
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return BigInt(cnt);
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}
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if (!limit && f[pos][cnt] !== -1n) {
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return f[pos][cnt];
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}
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let ans: bigint = 0n;
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let up: number = 1;
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if (limit) {
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up = Number((num >> BigInt(pos - 1)) & 1n);
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}
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for (let i = 0; i <= up; i++) {
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let v: number = cnt;
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if (i === 1 && pos % x === 0) {
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v++;
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}
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ans += dfs(pos - 1, v, limit && i === up);
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}
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if (!limit) {
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f[pos][cnt] = ans;
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}
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return ans;
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};
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while (l < r) {
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let mid: bigint = (l + r + 1n) >> 1n;
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num = mid;
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let m: number = num.toString(2).length;
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for (let i = 0; i < f.length; i++) {
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f[i].fill(-1n);
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}
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if (dfs(m, 0, true) <= BigInt(k)) {
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l = mid;
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} else {
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r = mid - 1n;
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}
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}
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return Number(l);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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@ -5,7 +5,7 @@ public:
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ll l = 1, r = 1e17;
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ll num = 0;
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ll f[65][65];
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function<ll(int, int, bool)> dfs = [&](int pos, int cnt, bool limit) -> ll {
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auto dfs = [&](auto&& dfs, int pos, int cnt, bool limit) -> ll {
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if (pos == 0) {
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return cnt;
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}
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@ -15,7 +15,7 @@ public:
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int up = limit ? num >> (pos - 1) & 1 : 1;
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ll ans = 0;
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for (int i = 0; i <= up; ++i) {
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ans += dfs(pos - 1, cnt + (i == 1 && pos % x == 0), limit && i == up);
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ans += dfs(dfs, pos - 1, cnt + (i == 1 && pos % x == 0), limit && i == up);
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}
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if (!limit) {
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f[pos][cnt] = ans;
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@ -27,7 +27,7 @@ public:
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num = mid;
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memset(f, -1, sizeof(f));
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int pos = 64 - __builtin_clzll(mid);
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if (dfs(pos, 0, true) <= k) {
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if (dfs(dfs, pos, 0, true) <= k) {
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l = mid;
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} else {
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r = mid - 1;
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@ -35,4 +35,4 @@ public:
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}
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return l;
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}
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};
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};
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@ -0,0 +1,45 @@
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function findMaximumNumber(k: number, x: number): number {
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let [l, r] = [1n, 10n ** 17n];
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let num: bigint;
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const f: bigint[][] = Array.from({ length: 65 }, () => Array(65).fill(-1n));
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const dfs = (pos: number, cnt: number, limit: boolean): bigint => {
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if (pos === 0) {
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return BigInt(cnt);
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}
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if (!limit && f[pos][cnt] !== -1n) {
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return f[pos][cnt];
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}
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let ans: bigint = 0n;
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let up: number = 1;
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if (limit) {
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up = Number((num >> BigInt(pos - 1)) & 1n);
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}
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for (let i = 0; i <= up; i++) {
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let v: number = cnt;
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if (i === 1 && pos % x === 0) {
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v++;
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}
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ans += dfs(pos - 1, v, limit && i === up);
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}
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if (!limit) {
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f[pos][cnt] = ans;
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}
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return ans;
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};
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while (l < r) {
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let mid: bigint = (l + r + 1n) >> 1n;
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num = mid;
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let m: number = num.toString(2).length;
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for (let i = 0; i < f.length; i++) {
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f[i].fill(-1n);
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}
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if (dfs(m, 0, true) <= BigInt(k)) {
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l = mid;
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} else {
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r = mid - 1n;
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}
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}
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return Number(l);
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}
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