2967. Minimum Cost to Make Array Equalindromic

Description

You are given a 0-indexed integer array nums having length n.

You are allowed to perform a special move any number of times (including zero) on nums. In one special move you perform the following steps in order:

Choose an index i in the range [0, n - 1], and a positive integer x.
Add |nums[i] - x| to the total cost.
Change the value of nums[i] to x.

A palindromic number is a positive integer that remains the same when its digits are reversed. For example, 121, 2552 and 65756 are palindromic numbers whereas 24, 46, 235 are not palindromic numbers.

An array is considered equalindromic if all the elements in the array are equal to an integer y, where y is a palindromic number less than 10⁹.

Return an integer denoting the minimum possible total cost to make nums equalindromic by performing any number of special moves.

Example 1:

Input: nums = [1,2,3,4,5]
Output: 6
Explanation: We can make the array equalindromic by changing all elements to 3 which is a palindromic number. The cost of changing the array to [3,3,3,3,3] using 4 special moves is given by |1 - 3| + |2 - 3| + |4 - 3| + |5 - 3| = 6.
It can be shown that changing all elements to any palindromic number other than 3 cannot be achieved at a lower cost.

Example 2:

Input: nums = [10,12,13,14,15]
Output: 11
Explanation: We can make the array equalindromic by changing all elements to 11 which is a palindromic number. The cost of changing the array to [11,11,11,11,11] using 5 special moves is given by |10 - 11| + |12 - 11| + |13 - 11| + |14 - 11| + |15 - 11| = 11.
It can be shown that changing all elements to any palindromic number other than 11 cannot be achieved at a lower cost.

Example 3:

Input: nums = [22,33,22,33,22]
Output: 22
Explanation: We can make the array equalindromic by changing all elements to 22 which is a palindromic number. The cost of changing the array to [22,22,22,22,22] using 2 special moves is given by |33 - 22| + |33 - 22| = 22.
It can be shown that changing all elements to any palindromic number other than 22 cannot be achieved at a lower cost.

Constraints:

1 <= n <= 10⁵
1 <= nums[i] <= 10⁹

Solutions

Solution 1: Preprocessing + Sorting + Binary Search

The range of palindrome numbers in the problem is $[1, 10^9]$. Due to the symmetry of palindrome numbers, we can enumerate in the range of $[1, 10^5]$, then reverse and concatenate them to get all palindrome numbers. Note that if it is an odd-length palindrome number, we need to remove the last digit before reversing. The array of palindrome numbers obtained by preprocessing is denoted as $ps$. We sort the array $ps$.

Next, we sort the array $nums$ and take the median $x$ of $nums$. We only need to find a number in the palindrome array $ps$ that is closest to $x$ through binary search, and then calculate the cost of $nums$ becoming this number to get the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array $nums$, and $M$ is the length of the palindrome array $ps$.

public class Solution {
    private static long[] ps;
    private int[] nums;

    static {
        ps = new long[2 * (int) 1e5];
        for (int i = 1; i <= 1e5; i++) {
            String s = Integer.toString(i);
            String t1 = new StringBuilder(s).reverse().toString();
            String t2 = new StringBuilder(s.substring(0, s.length() - 1)).reverse().toString();
            ps[2 * i - 2] = Long.parseLong(s + t1);
            ps[2 * i - 1] = Long.parseLong(s + t2);
        }
        Arrays.sort(ps);
    }

    public long minimumCost(int[] nums) {
        this.nums = nums;
        Arrays.sort(nums);
        int i = Arrays.binarySearch(ps, nums[nums.length / 2]);
        i = i < 0 ? -i - 1 : i;
        long ans = 1L << 60;
        for (int j = i - 1; j <= i + 1; j++) {
            if (0 <= j && j < ps.length) {
                ans = Math.min(ans, f(ps[j]));
            }
        }
        return ans;
    }

    private long f(long x) {
        long ans = 0;
        for (int v : nums) {
            ans += Math.abs(v - x);
        }
        return ans;
    }
}

using ll = long long;

ll ps[2 * 100000];

int init = [] {
    for (int i = 1; i <= 100000; i++) {
        string s = to_string(i);
        string t1 = s;
        reverse(t1.begin(), t1.end());
        string t2 = s.substr(0, s.length() - 1);
        reverse(t2.begin(), t2.end());
        ps[2 * i - 2] = stoll(s + t1);
        ps[2 * i - 1] = stoll(s + t2);
    }
    sort(ps, ps + 2 * 100000);
    return 0;
}();

class Solution {
public:
    long long minimumCost(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int i = lower_bound(ps, ps + 2 * 100000, nums[nums.size() / 2]) - ps;
        auto f = [&](ll x) {
            ll ans = 0;
            for (int& v : nums) {
                ans += abs(v - x);
            }
            return ans;
        };
        ll ans = LLONG_MAX;
        for (int j = i - 1; j <= i + 1; j++) {
            if (0 <= j && j < 2 * 100000) {
                ans = min(ans, f(ps[j]));
            }
        }
        return ans;
    }
};

ps = []
for i in range(1, 10**5 + 1):
    s = str(i)
    t1 = s[::-1]
    t2 = s[:-1][::-1]
    ps.append(int(s + t1))
    ps.append(int(s + t2))
ps.sort()


class Solution:
    def minimumCost(self, nums: List[int]) -> int:
        def f(x: int) -> int:
            return sum(abs(v - x) for v in nums)

        nums.sort()
        i = bisect_left(ps, nums[len(nums) // 2])
        return min(f(ps[j]) for j in range(i - 1, i + 2) if 0 <= j < len(ps))

var ps [2 * 100000]int64

func init() {
	for i := 1; i <= 100000; i++ {
		s := strconv.Itoa(i)
		t1 := reverseString(s)
		t2 := reverseString(s[:len(s)-1])
		ps[2*i-2], _ = strconv.ParseInt(s+t1, 10, 64)
		ps[2*i-1], _ = strconv.ParseInt(s+t2, 10, 64)
	}
	sort.Slice(ps[:], func(i, j int) bool {
		return ps[i] < ps[j]
	})
}

func reverseString(s string) string {
	cs := []rune(s)
	for i, j := 0, len(cs)-1; i < j; i, j = i+1, j-1 {
		cs[i], cs[j] = cs[j], cs[i]
	}
	return string(cs)
}

func minimumCost(nums []int) int64 {
	sort.Ints(nums)
	i := sort.Search(len(ps), func(i int) bool {
		return ps[i] >= int64(nums[len(nums)/2])
	})

	f := func(x int64) int64 {
		var ans int64
		for _, v := range nums {
			ans += int64(abs(int(x - int64(v))))
		}
		return ans
	}

	ans := int64(math.MaxInt64)
	for j := i - 1; j <= i+1; j++ {
		if 0 <= j && j < len(ps) {
			ans = min(ans, f(ps[j]))
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

const ps = Array(2e5).fill(0);

const init = (() => {
    for (let i = 1; i <= 1e5; ++i) {
        const s: string = i.toString();
        const t1: string = s.split('').reverse().join('');
        const t2: string = s.slice(0, -1).split('').reverse().join('');
        ps[2 * i - 2] = parseInt(s + t1, 10);
        ps[2 * i - 1] = parseInt(s + t2, 10);
    }
    ps.sort((a, b) => a - b);
})();

function minimumCost(nums: number[]): number {
    const search = (x: number): number => {
        let [l, r] = [0, ps.length];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (ps[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const f = (x: number): number => {
        return nums.reduce((acc, v) => acc + Math.abs(v - x), 0);
    };

    nums.sort((a, b) => a - b);
    const i: number = search(nums[nums.length >> 1]);
    let ans: number = Number.MAX_SAFE_INTEGER;
    for (let j = i - 1; j <= i + 1; j++) {
        if (j >= 0 && j < ps.length) {
            ans = Math.min(ans, f(ps[j]));
        }
    }
    return ans;
}

2967 - Minimum Cost to Make Array Equalindromic

2967. Minimum Cost to Make Array Equalindromic

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2967. Minimum Cost to Make Array Equalindromic

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