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authorJack Sangdahl <[email protected]>2025-04-07 21:15:38 -0600
committerJack Sangdahl <[email protected]>2025-04-07 21:15:48 -0600
commit3fdd56c0adaccc960214af925cdc5c19bf3d187a (patch)
tree528b37c78a834e01983ca893b96b8be6e833d6a6 /min_num_op_elem_distinct.c
parent6c6072834b6854644ce87c4182a069d2f7358da7 (diff)
downloadleetcode-3fdd56c0adaccc960214af925cdc5c19bf3d187a.tar.gz
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3396. minimum number of operations to make elements in array distinct
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+// 3396. Minimum Number of Operations to Make Elements in Array Distinct
+#include "leetcode.h"
+
+/*
+ * you are given an integer array 'nums'. you need to ensure that the elements
+ * in the array are distinct. to achieve this, you can remove 3 elements from
+ * the beginning of the array. if the array has fewer than 3 elements, remove
+ * all remaining elements. note that an empty array is considered to have
+ * distinct elements. return the minimum number of operations needed to make the
+ * elements in the array distinct.
+ */
+
+int minimumOperations(int *nums, int numsSize) {
+ char prev[101], max = -1;
+ memset(prev, -1, sizeof(prev));
+ for (int i = 0; i < numsSize; i++) {
+ if (prev[nums[i]] != -1 && max < prev[nums[i]])
+ max = prev[nums[i]];
+ prev[nums[i]] = i;
+ }
+ return (int)(max + 1) / 3 + ((max + 1) % 3 > 0);
+}
+
+int main() {
+ int n1[] = {1, 2, 3, 4, 2, 3, 3, 5, 7}, n2[] = {4, 5, 6, 4, 4},
+ n3[] = {6, 7, 8, 9};
+ printf("%d\n", minimumOperations(n1, ARRAY_SIZE(n1))); // expect: 2
+ printf("%d\n", minimumOperations(n2, ARRAY_SIZE(n2))); // expect: 2
+ printf("%d\n", minimumOperations(n3, ARRAY_SIZE(n3))); // expect: 0
+}