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2460.apply-operations-to-an-array.cpp
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65 lines (63 loc) · 2.22 KB
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// Tag: Array, Two Pointers, Simulation
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// Video: https://youtu.be/l-KC4smk5sM
// You are given a 0-indexed array nums of size n consisting of non-negative integers.
// You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:
//
// If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.
//
// After performing all the operations, shift all the 0's to the end of the array.
//
// For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0].
//
// Return the resulting array.
// Note that the operations are applied sequentially, not all at once.
//
// Example 1:
//
// Input: nums = [1,2,2,1,1,0]
// Output: [1,4,2,0,0,0]
// Explanation: We do the following operations:
// - i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
// - i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
// - i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
// - i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
// - i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
// After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].
//
// Example 2:
//
// Input: nums = [0,1]
// Output: [1,0]
// Explanation: No operation can be applied, we just shift the 0 to the end.
//
//
// Constraints:
//
// 2 <= nums.length <= 2000
// 0 <= nums[i] <= 1000
//
//
class Solution {
public:
vector<int> applyOperations(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n - 1; i++) {
if (nums[i] == nums[i + 1]) {
nums[i] *= 2;
nums[i + 1] = 0;
}
}
int j = 0;
for (int i = 0; i < n; i++) {
if (nums[i]) {
swap(nums[i], nums[j]);
j += 1;
}
}
return nums;
}
};