2163. Minimum Difference in Sums After Removal of Elements: RETRY

Author: tan-wei

src/solution/mod.rs

@@ -1634,3 +1634,4 @@ mod s2157_groups_of_strings;
 mod s2160_minimum_sum_of_four_digit_number_after_splitting_digits;
 mod s2161_partition_array_according_to_given_pivot;
 mod s2162_minimum_cost_to_set_cooking_time;
+mod s2163_minimum_difference_in_sums_after_removal_of_elements;

src/solution/s2163_minimum_difference_in_sums_after_removal_of_elements.rs

@@ -0,0 +1,83 @@
+/**
+ * [2163] Minimum Difference in Sums After Removal of Elements
+ *
+ * You are given a 0-indexed integer array nums consisting of 3 * n elements.
+ * You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:
+ *
+ * 	The first n elements belonging to the first part and their sum is sumfirst.
+ * 	The next n elements belonging to the second part and their sum is sumsecond.
+ *
+ * The difference in sums of the two parts is denoted as sumfirst - sumsecond.
+ *
+ * 	For example, if sumfirst = 3 and sumsecond = 2, their difference is 1.
+ * 	Similarly, if sumfirst = 2 and sumsecond = 3, their difference is -1.
+ *
+ * Return the minimum difference possible between the sums of the two parts after the removal of n elements.
+ *  
+ * Example 1:
+ *
+ * Input: nums = [3,1,2]
+ * Output: -1
+ * Explanation: Here, nums has 3 elements, so n = 1.
+ * Thus we have to remove 1 element from nums and divide the array into two equal parts.
+ * - If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.
+ * - If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.
+ * - If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.
+ * The minimum difference between sums of the two parts is min(-1,1,2) = -1.
+ *
+ * Example 2:
+ *
+ * Input: nums = [7,9,5,8,1,3]
+ * Output: 1
+ * Explanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.
+ * If we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.
+ * To obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.
+ * It can be shown that it is not possible to obtain a difference smaller than 1.
+ *
+ *  
+ * Constraints:
+ *
+ * 	nums.length == 3 * n
+ * 	1 <= n <= 10^5
+ * 	1 <= nums[i] <= 10^5
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/minimum-difference-in-sums-after-removal-of-elements/
+// discuss: https://leetcode.com/problems/minimum-difference-in-sums-after-removal-of-elements/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn minimum_difference(nums: Vec<i32>) -> i64 {
+        0
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    #[ignore]
+    fn test_2163_example_1() {
+        let nums = vec![3, 1, 2];
+
+        let result = -1;
+
+        assert_eq!(Solution::minimum_difference(nums), result);
+    }
+
+    #[test]
+    #[ignore]
+    fn test_2163_example_2() {
+        let nums = vec![7, 9, 5, 8, 1, 3];
+
+        let result = 1;
+
+        assert_eq!(Solution::minimum_difference(nums), result);
+    }
+}