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Added task 698.

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	`1`	`+package g0601_0700.s0698_partition_to_k_equal_sum_subsets;`
	`2`	`+`
	`3`	`+// #Medium #Array #Dynamic_Programming #Bit_Manipulation #Backtracking #Bitmask #Memoization`
	`4`	`+`
	`5`	`+import java.util.Arrays;`
	`6`	`+`
	`7`	`+public class Solution {`
	`8`	`+ public boolean canPartitionKSubsets(int[] nums, int k) {`
	`9`	`+ if (nums == null \|\| nums.length == 0) {`
	`10`	`+ return false;`
	`11`	`+ }`
	`12`	`+ int n = nums.length;`
	`13`	`+ int sum = 0;`
	`14`	`+ for (int num : nums) {`
	`15`	`+ sum += num;`
	`16`	`+ }`
	`17`	`+ if (sum % k != 0) {`
	`18`	`+ return false;`
	`19`	`+ }`
	`20`	`+ // sum of each subset = sum / k`
	`21`	`+ sum /= k;`
	`22`	`+ int[] dp = new int[1 << n];`
	`23`	`+ Arrays.fill(dp, -1);`
	`24`	`+ dp[0] = 0;`
	`25`	`+ for (int i = 0; i < (1 << n); i++) {`
	`26`	`+ if (dp[i] == -1) {`
	`27`	`+ continue;`
	`28`	`+ }`
	`29`	`+ int rem = sum - (dp[i] % sum);`
	`30`	`+ for (int j = 0; j < n; j++) {`
	`31`	`+ // bitmask`
	`32`	`+ int tmp = i \| (1 << j);`
	`33`	`+ // skip if the bit is already taken`
	`34`	`+ if (tmp != i) {`
	`35`	`+ // num too big for current subset`
	`36`	`+ if (nums[j] > rem) {`
	`37`	`+ break;`
	`38`	`+ }`
	`39`	`+ // cumulative sum`
	`40`	`+ dp[tmp] = dp[i] + nums[j];`
	`41`	`+ }`
	`42`	`+ }`
	`43`	`+ }`
	`44`	`+ // true if total sum of all nums is the same`
	`45`	`+ return dp[(1 << n) - 1] == k * sum;`
	`46`	`+ }`
	`47`	`+}`

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	`1`	`+698\. Partition to K Equal Sum Subsets`
	`2`	`+`
	`3`	`+Medium`
	`4`	`+`
	`5`	+Given an integer array `nums` and an integer `k`, return `true` if it is possible to divide this array into `k` non-empty subsets whose sums are all equal.
	`6`	`+`
	`7`	`+Example 1:`
	`8`	`+`
	`9`	`+Input: nums = [4,3,2,3,5,2,1], k = 4`
	`10`	`+`
	`11`	`+Output: true`
	`12`	`+`
	`13`	`+Explanation: It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.`
	`14`	`+`
	`15`	`+Example 2:`
	`16`	`+`
	`17`	`+Input: nums = [1,2,3,4], k = 3`
	`18`	`+`
	`19`	`+Output: false`
	`20`	`+`
	`21`	`+Constraints:`
	`22`	`+`
	`23`	+* `1 <= k <= nums.length <= 16`
	`24`	`+* <code>1 <= nums[i] <= 10<sup>4</sup></code>`
	`25`	+* The frequency of each element is in the range `[1, 4]`.

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	`1`	`+package g0601_0700.s0698_partition_to_k_equal_sum_subsets;`
	`2`	`+`
	`3`	`+import static org.hamcrest.CoreMatchers.equalTo;`
	`4`	`+import static org.hamcrest.MatcherAssert.assertThat;`
	`5`	`+`
	`6`	`+import org.junit.jupiter.api.Test;`
	`7`	`+`
	`8`	`+class SolutionTest {`
	`9`	`+ @Test`
	`10`	`+ void canPartitionKSubsets() {`
	`11`	`+ assertThat(`
	`12`	`+ new Solution().canPartitionKSubsets(new int[] {4, 3, 2, 3, 5, 2, 1}, 4),`
	`13`	`+ equalTo(true));`
	`14`	`+ }`
	`15`	`+`
	`16`	`+ @Test`
	`17`	`+ void canPartitionKSubsets2() {`
	`18`	`+ assertThat(new Solution().canPartitionKSubsets(new int[] {1, 2, 3, 4}, 3), equalTo(false));`
	`19`	`+ }`
	`20`	`+}`

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