feat: add solutions for lc No.3888 by yanglbme · Pull Request #5128 · doocs/leetcode

yanglbme · 2026-04-01T00:43:55Z

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Copilot

Pull request overview

Adds repository documentation and multi-language reference implementations for LeetCode problem 3888 (locked), and registers the problem in the global solution indexes.

Changes:

Added solution index entries for problem 3888 in solution/README.md and solution/README_EN.md.
Added new problem documentation (README.md, README_EN.md) and implementations (C++/Go/Java/Python/TypeScript) under solution/3800-3899/3888....
Minor formatting change in an existing locked problem English README (2753).

Reviewed changes

Copilot reviewed 9 out of 11 changed files in this pull request and generated 9 comments.

Show a summary per file

File	Description
solution/README.md	Adds problem 3888 entry to the Chinese solutions index.
solution/README_EN.md	Adds problem 3888 entry to the English solutions index.
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/README.md	Adds Chinese problem README + solution explanation (but currently contains correctness/consistency issues).
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/README_EN.md	Adds English problem README + solution explanation (but currently contains correctness issues).
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/Solution.ts	Adds TypeScript implementation (target-selection logic appears incorrect).
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/Solution.py	Adds Python implementation (target-selection logic appears incorrect).
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/Solution.java	Adds Java implementation (target-selection logic appears incorrect).
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/Solution.go	Adds Go implementation (target-selection logic appears incorrect).
solution/3800-3899/3888.Minimum Operations to Make All Grid Elements Equal/Solution.cpp	Adds C++ implementation (target-selection logic appears incorrect).
solution/2700-2799/2753.Count Houses in a Circular Street II/README_EN.md	Minor formatting/whitespace change.

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Copilot · 2026-04-01T00:53:17Z

+    for (let t = maxVal; t <= maxVal + 1; t++) {
+        const res = check(t);
+        if (res !== -1) return res;
+    }
+
+    return -1;


The algorithm only tries target values max(grid) and max(grid)+1, but a valid solution may require a larger target. For example with k=2 and grid [[900,700,800],[600,0,400],[700,300,600]], the unique feasible target is 1000 (so this code would incorrectly return -1). The target selection logic needs to be generalized; do not assume only these two candidates.

Suggested change

for (let t = maxVal; t <= maxVal + 1; t++) {

const res = check(t);

if (res !== -1) return res;

}

return -1;

let left = maxVal;

let right = maxVal + 1_000_000_000;

let answer = -1;

while (left <= right) {

const mid = Math.floor((left + right) / 2);

const res = check(mid);

if (res !== -1) {

answer = res;

right = mid - 1;

} else {

left = mid + 1;

}

}

return answer;

Copilot · 2026-04-01T00:53:18Z

+        for t in range(mx, mx + 2):
+            res = check(t)
+            if res != -1:
+                return res
+
+        return -1


This implementation only checks target values in {max(grid), max(grid)+1}. That assumption is false in general for k×k increments: there are grids where the only feasible equal value is > max(grid)+1, so this can return -1 even when a solution exists. Please replace the target selection with a correct method (and update the explanation in the README accordingly).

Copilot · 2026-04-01T00:53:18Z

+        for (int t = mx; t <= mx + 1; t++) {
+            long res = check(t);
+            if (res != -1) {
+                return res;
+            }
+        }
+        return -1;


The loop only tests target values mx and mx+1, but feasible instances can require a larger target (so this can incorrectly return -1). The solution needs a correct way to determine/derive the required final value instead of assuming it is within +1 of the initial maximum.

Suggested change

for (int t = mx; t <= mx + 1; t++) {

long res = check(t);

if (res != -1) {

return res;

}

}

return -1;

// We need to find the minimal feasible target T >= mx such that check(T) != -1.

// Feasibility is monotone in T: once a T is feasible, all larger T are feasible as well.

long lo = mx;

long hi = mx;

long bestOps = -1;

// Phase 1: exponentially increase hi until we find a feasible target or hit a safe cap.

while (true) {

if (hi > Integer.MAX_VALUE) {

hi = Integer.MAX_VALUE;

}

long res = check((int) hi);

if (res != -1) {

bestOps = res;

break;

}

if (hi == Integer.MAX_VALUE) {

// Even the largest possible int target is not feasible.

return -1;

}

hi <<= 1;

}

// Phase 2: binary search between lo and hi for the minimal feasible target.

long left = lo;

long right = hi;

while (left <= right) {

long mid = left + ((right - left) >>> 1);

if (mid > Integer.MAX_VALUE) {

// Targets beyond Integer.MAX_VALUE are irrelevant since check takes an int.

right = Integer.MAX_VALUE;

continue;

}

long res = check((int) mid);

if (res != -1) {

bestOps = res;

right = mid - 1;

} else {

left = mid + 1;

}

}

return bestOps;

Copilot · 2026-04-01T00:53:19Z

+	for t := maxVal; t <= maxVal+1; t++ {
+		if res := check(t); res != -1 {
+			return res
+		}
+	}
+
+	return -1


This code assumes the final equal value must be maxVal or maxVal+1 by only calling check on those two targets. That is not generally true for k×k increment operations; some solvable grids require a target > maxVal+1, so this would incorrectly return -1. Target selection should be reworked to handle larger required targets.

Copilot · 2026-04-01T00:53:19Z

+        for (int t = mx; t <= mx + 1; ++t) {
+            long long res = check(t);
+            if (res != -1) {
+                return res;
+            }
+        }
+
+        return -1;
+    }


The solution only considers targets t in [mx, mx+1], but solvable cases exist where the required target is greater than mx+1. This target restriction can cause false negatives (-1). Please remove this assumption and implement a correct target derivation/validation strategy.

Copilot · 2026-04-01T00:53:19Z

+由于操作只能增加元素的值，因此最终网格的所有元素必须等于某个目标值 $T$，且 $T \ge \max(\textit{grid})$。
+
+从左上角 $(0, 0)$ 开始遍历网格。对于任意位置 $(i, j)$，如果它当前的数值小于 $T$，由于后续的操作（以更靠右或更靠下的位置为左上角的操作）都无法覆盖到 $(i, j)$，因此必须在当前位置执行 $T - \text{current\_val}$ 次以 $(i, j)$ 为左上角的 $k \times k$ 增加操作。
+
+如果每次执行操作都遍历 $k \times k$ 区域，复杂度将达到 $O(m \cdot n \cdot k^2)$。我们可以使用二维差分数组 $\textit{diff}$ 来记录操作。通过实时维护 $\textit{diff}$ 的二维前缀和，我们可以在 $O(1)$ 时间内获取当前位置的累计增量，并在 $O(1)$ 时间内更新一个 $k \times k$ 区域的未来影响。
+
+通常情况下 $T = \max(\textit{grid})$ 即可。但在某些 $k \times k$ 覆盖重叠的情况下，较小的 $T$ 可能导致中间位置被动增加后超过 $T$。根据数学一致性，若 $T = \max(\textit{grid})$ 和 $T = \max(\textit{grid}) + 1$ 均不可行，则该网格无法通过 $k \times k$ 操作变平。
+
+时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是网格的行数和列数。


The explanation claims that only T = max(grid) or max(grid)+1 need to be checked, and concludes the grid is impossible otherwise. This claim is not true in general for k×k increment operations; there are solvable grids where the only feasible target is > max(grid)+1. The write-up (and the implementations) should be updated to use a correct target determination approach.

Copilot · 2026-04-01T00:53:19Z

+Since the operation can only increase the value of elements, all elements in the final grid must be equal to some target value $T$, and $T \ge \max(\textit{grid})$.
+
+Start traversing the grid from the top-left corner $(0, 0)$. For any position $(i, j)$, if its current value is less than $T$, since subsequent operations (with a more rightward or downward position as the top-left corner) cannot cover $(i, j)$, it is necessary to perform $T - \text{current\_val}$ operations at the current position, each using $(i, j)$ as the top-left corner of a $k \times k$ increment operation.
+
+If each operation traverses the $k \times k$ region, the complexity will reach $O(m \cdot n \cdot k^2)$. We can use a 2D difference array $\textit{diff}$ to record the operations. By maintaining the 2D prefix sum of $\textit{diff}$ in real time, we can obtain the cumulative increment at the current position in $O(1)$ time, and update the future impact of a $k \times k$ region in $O(1)$ time.
+
+In most cases, $T = \max(\textit{grid})$ is sufficient. However, in some cases where $k \times k$ regions overlap, a smaller $T$ may cause the middle positions to be passively increased beyond $T$. According to mathematical consistency, if both $T = \max(\textit{grid})$ and $T = \max(\textit{grid}) + 1$ are not feasible, then it is impossible to flatten the grid using $k \times k$ operations.
+
+The time complexity is $O(m \times n)$ and the space complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the grid, respectively.


The explanation states that checking only T = max(grid) and T = max(grid)+1 is sufficient. That assumption can be false: some instances require a larger target, so the described approach can incorrectly return -1. Please update the explanation to match a corrected algorithm for determining the final target value.

Copilot · 2026-04-01T00:53:19Z

+# [3888. Minimum Operations to Make All Grid Elements Equal 🔒](https://leetcode.cn/problems/minimum-operations-to-make-all-grid-elements-equal)
+
+[English Version](/solution/3800-3899/3888.Minimum%20Operations%20to%20Make%20All%20Grid%20Elements%20Equal/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>You are given a 2D integer array <code>grid</code> of size <code>m &times; n</code>, and an integer <code>k</code>.</p>
+
+<p>In one operation, you can:</p>
+
+<ul>
+	<li>Select any <code>k x k</code> <strong>submatrix</strong> of <code>grid</code>, and</li>
+	<li>Increment <strong>all elements</strong> inside that <strong>submatrix</strong> by 1.</li>
+</ul>
+
+<p>Return the <strong>minimum</strong> number of operations required to make all elements in the grid <strong>equal</strong>. If it is not possible, return -1.</p>
+A submatrix <code>(x1, y1, x2, y2)</code> is a matrix that forms by choosing all cells <code>matrix[x][y]</code> where <code>x1 &lt;= x &lt;= x2</code> and <code>y1 &lt;= y &lt;= y2</code>.


This Chinese README uses an English problem title and an untranslated English statement under "题目描述", which is inconsistent with other Chinese READMEs in this repo (including locked problems) that provide Chinese titles/descriptions. Please translate/localize the title and statement for consistency.

Copilot · 2026-04-01T00:53:20Z

 <strong>Input:</strong> street = [1,1,1,1], k = 10
 <strong>Output:</strong> 4
-<strong>Explanation:</strong> There are 4 houses, and all their doors are open.
+<strong>Explanation:</strong> There are 4 houses, and all their doors are open. 


There is trailing whitespace at the end of this line (after the period). Please remove it to avoid noisy diffs and keep formatting consistent.

Suggested change

<strong>Explanation:</strong> There are 4 houses, and all their doors are open.

<strong>Explanation:</strong> There are 4 houses, and all their doors are open.

feat: add solutions for lc No.3888

3a0eb15

Copilot AI review requested due to automatic review settings April 1, 2026 00:43

yanglbme merged commit e9cfab5 into main Apr 1, 2026
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yanglbme deleted the dev branch April 1, 2026 00:44

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feat: add solutions for lc No.3888#5128

feat: add solutions for lc No.3888#5128
yanglbme merged 1 commit intomainfrom
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-    for (let t = maxVal; t <= maxVal + 1; t++) {
-        const res = check(t);
-        if (res !== -1) return res;
-    }
-    return -1;
+    let left = maxVal;
+    let right = maxVal + 1_000_000_000;
+    let answer = -1;
+    while (left <= right) {
+        const mid = Math.floor((left + right) / 2);
+        const res = check(mid);
+        if (res !== -1) {
+            answer = res;
+            right = mid - 1;
+        } else {
+            left = mid + 1;
+        }
+    }
+    return answer;

-        for (int t = mx; t <= mx + 1; t++) {
-            long res = check(t);
-            if (res != -1) {
-                return res;
-            }
-        }
-        return -1;
+        // We need to find the minimal feasible target T >= mx such that check(T) != -1.
+        // Feasibility is monotone in T: once a T is feasible, all larger T are feasible as well.
+        long lo = mx;
+        long hi = mx;
+        long bestOps = -1;
+        // Phase 1: exponentially increase hi until we find a feasible target or hit a safe cap.
+        while (true) {
+            if (hi > Integer.MAX_VALUE) {
+                hi = Integer.MAX_VALUE;
+            }
+            long res = check((int) hi);
+            if (res != -1) {
+                bestOps = res;
+                break;
+            }
+            if (hi == Integer.MAX_VALUE) {
+                // Even the largest possible int target is not feasible.
+                return -1;
+            }
+            hi <<= 1;
+        }
+        // Phase 2: binary search between lo and hi for the minimal feasible target.
+        long left = lo;
+        long right = hi;
+        while (left <= right) {
+            long mid = left + ((right - left) >>> 1);
+            if (mid > Integer.MAX_VALUE) {
+                // Targets beyond Integer.MAX_VALUE are irrelevant since check takes an int.
+                right = Integer.MAX_VALUE;
+                continue;
+            }
+            long res = check((int) mid);
+            if (res != -1) {
+                bestOps = res;
+                right = mid - 1;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return bestOps;

	<strong>Explanation:</strong> There are 4 houses, and all their doors are open.
	<strong>Explanation:</strong> There are 4 houses, and all their doors are open.

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