LeetCode #2967 — MEDIUM

Minimum Cost to Make Array Equalindromic

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Binary Search · Greedy

Example 1

[1,2,3,4,5]

Example 2

[10,12,13,14,15]

Example 3

[22,33,22,33,22]

Core Insight

What unlocks the optimal approach

Find the median of <code>nums</code> after sorting it (if the length is even, we can select any number from the two in the middle). Let’s call it <code>m</code>.
Try the smallest palindromic number that is larger than or equal to <code>m</code> (if any) and the largest palindromic number that is smaller than or equal to <code>m</code> (if any). These two values are the candidate palindromic numbers for values of all indices.
We can use math constructions to construct the two palindromic numbers in <code>O(log(m) / 2)</code> time or we can do it using brute-force by starting from m and checking smaller and larger values in <code>O(sqrt(10log(m)))</code>.
It is also possible to just generate all palindromic numbers using recursion in <code>O(sqrt(109log(109))</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2967: Minimum Cost to Make Array Equalindromic
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;
    }
}

// Accepted solution for LeetCode #2967: Minimum Cost to Make Array Equalindromic
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
}

# Accepted solution for LeetCode #2967: Minimum Cost to Make Array Equalindromic
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))

// Accepted solution for LeetCode #2967: Minimum Cost to Make Array Equalindromic
impl Solution {
    pub fn minimum_cost(nums: Vec<i32>) -> i64 {
        use std::cmp::min;
        use std::sync::Once;

        static INIT: Once = Once::new();
        static mut PS: Vec<i64> = Vec::new();

        INIT.call_once(|| {
            let mut ps_local = Vec::with_capacity(2 * 100_000);
            for i in 1..=100_000 {
                let s = i.to_string();

                let mut t1 = s.clone();
                t1 = t1.chars().rev().collect();
                ps_local.push(format!("{}{}", s, t1).parse::<i64>().unwrap());

                let mut t2 = s[0..s.len() - 1].to_string();
                t2 = t2.chars().rev().collect();
                ps_local.push(format!("{}{}", s, t2).parse::<i64>().unwrap());
            }
            ps_local.sort();
            unsafe {
                PS = ps_local;
            }
        });

        let mut nums = nums;
        nums.sort();

        let mid = nums[nums.len() / 2] as i64;

        let i = unsafe {
            match PS.binary_search(&mid) {
                Ok(i) => i,
                Err(i) => i,
            }
        };

        let f = |x: i64| -> i64 { nums.iter().map(|&v| (v as i64 - x).abs()).sum() };

        let mut ans = i64::MAX;

        for j in i.saturating_sub(1)..=(i + 1).min(2 * 100_000 - 1) {
            let x = unsafe { PS[j] };
            ans = min(ans, f(x));
        }

        ans
    }
}

// Accepted solution for LeetCode #2967: Minimum Cost to Make Array Equalindromic
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(M)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.