--- title: Cut Matrix: A Deep Dive into the DP + Binary Search Solution description: A detailed walkthrough of the Cut Matrix problem, where a binary grid must be sliced into exactly k rectangular pieces each containing at least one 1, using a... canonical: https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx twitter:card: summary_large_image twitter:site: @dailydotdev og:type: website og:site_name: daily.dev og:title: Cut Matrix: A Deep Dive into the DP + Binary Search Solution | daily.dev og:description: A detailed walkthrough of the Cut Matrix problem, where a binary grid must be sliced into exactly k rectangular pieces each containing at least one 1, using a... og:url: https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx og:image: https://api.daily.dev/og/posts/WMRbvF8LX.png og:image:alt: Cut Matrix: A Deep Dive into the DP + Binary Search Solution og:image:width: 1200 og:image:height: 630 og:locale: en --- > ## Documentation Index > Fetch the complete documentation index at: https://daily.dev/llms.txt > Use this file to discover all available pages before exploring further. # Cut Matrix: A Deep Dive into the DP + Binary Search Solution **[C\# Corner](https://daily.dev/sources/csharpcorner)** · 6 min read · 1 upvotes · 0 comments ## Summary A detailed walkthrough of the Cut Matrix problem, where a binary grid must be sliced into exactly k rectangular pieces each containing at least one 1, using a constrained cutting rule (only the bottom or right remainder stays cuttable). The key insight is that the active region is always a suffix rectangle defined by just two coordinates, reducing the state space to O(n×m). The solution combines 2D prefix sums for O(1) validity checks, binary search exploiting monotonicity to find the first valid cut boundary, and suffix sums of the DP layer below to aggregate cut contributions in O(1). The resulting algorithm runs in O(n×m×k×log(n+m)) time with a full Java implementation provided. ## Full article daily.dev links to this article rather than hosting it. Read it at the original source: ## Questions this post answers ### How do you reduce the state space for a matrix-cutting DP problem where each cut only leaves the bottom or right piece cuttable? The cutting rule restricts the remaining cuttable region to always be a suffix rectangle described by just a starting row r and starting column c, extending to the matrix's bottom-right corner. Because horizontal cuts only shrink the top boundary and vertical cuts only shrink the left boundary, the state space collapses from arbitrary rectangles to O(n*m) possible active regions. _daily.dev surfaces algorithmic deep dives like this for engineers sharpening DP and binary search skills._ ### How can binary search speed up a DP recurrence that checks whether a matrix piece contains at least one 1? As the cut position moves farther from the anchor, the peeled-off piece only grows, so the count of 1s inside it is monotonically non-decreasing, which lets you binary search for the first valid cut index instead of scanning every position. Combined with a 2D prefix-sum array giving O(1) piece checks, this cuts per-state cost from O(n+m) to O(log(n+m)). _Engineers optimizing DP solutions can find similar technique breakdowns curated on daily.dev._ ### What is the time and space complexity of the DP plus binary search plus suffix sum solution for counting valid ways to cut a matrix into k pieces? The solution runs in O(n * m * k * log(n + m)) time and O(n * m * k) space. This comes from k DP layers, each covering n*m cells, with two binary searches per cell, plus a full 3D DP table alongside O(n*m) suffix-sum scratch space reused per layer. _daily.dev helps developers track complexity-focused algorithm writeups when comparing solution approaches._ ## Similar posts on daily.dev - [Blind 75 \| Two Pointers — 15 Spiral Matrix](https://daily.dev/posts/blind-75-two-pointers-15-spiral-matrix-1mx7uxr6i) · Medium · 0 upvotes · 0 comments - [Count Substrings with More 1's than 0's](https://daily.dev/posts/count-substrings-with-more-1-s-than-0-s-k63z1kpmf) · C\# Corner · 0 upvotes · 0 comments - [The Subset Sum Problem Solved in Linear Time for Dense Enough Inputs](https://daily.dev/posts/the-subset-sum-problem-solved-in-linear-time-for-dense-enough-inputs-sjlhgprsn) · Towards Data Science · 0 upvotes · 0 comments - [Spatial DP: Finding the Largest Square — Monday Morning Haskell](https://daily.dev/posts/spatial-dp-finding-the-largest-square-monday-morning-haskell-owyi7epbd) · Monday Morning Haskell · 0 upvotes · 0 comments --- Tags: [#java](https://daily.dev/tags/java), [#algorithms](https://daily.dev/tags/algorithms), [#binary-search](https://daily.dev/tags/binary-search), [#dynamic-programming](https://daily.dev/tags/dynamic-programming) [View this post on daily.dev](https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx) ```json {"@context":"https://schema.org","@graph":[{"@type":"Organization","@id":"https://daily.dev/#organization","name":"daily.dev","url":"https://daily.dev","logo":{"@type":"ImageObject","url":"https://daily.dev/apple-touch-icon.png","width":180,"height":180},"sameAs":["https://twitter.com/dailydotdev","https://github.com/dailydotdev","https://www.linkedin.com/company/daily-dev-ltd"]},{"@type":"WebSite","@id":"https://daily.dev/#website","url":"https://daily.dev","name":"daily.dev","publisher":{"@id":"https://daily.dev/#organization"},"potentialAction":{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https://daily.dev/search?q={search_term_string}"},"query-input":"required name=search_term_string"}}]} {"@context":"https://schema.org","@type":"TechArticle","headline":"Cut Matrix: A Deep Dive into the DP + Binary Search Solution","url":"https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx","mainEntityOfPage":{"@type":"WebPage","@id":"https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx"},"datePublished":"2026-07-20T06:31:48.511Z","dateModified":"2026-09-14T09:28:37.344Z","description":"A detailed walkthrough of the Cut Matrix problem, where a binary grid must be sliced into exactly k rectangular pieces each containing at least one 1, using a...","image":"https://media.daily.dev/image/upload/f_auto,q_auto/v1/posts/81f698b062e81502c45df6b257ae3843?_a=AQAEuop","thumbnailUrl":"https://media.daily.dev/image/upload/f_auto,q_auto/v1/posts/81f698b062e81502c45df6b257ae3843?_a=AQAEuop","isAccessibleForFree":true,"articleSection":"C# Corner","inLanguage":"en","publisher":{"@type":"Organization","name":"daily.dev","url":"https://daily.dev","logo":{"@type":"ImageObject","url":"https://daily.dev/apple-touch-icon.png","width":180,"height":180}},"author":{"@type":"Organization","name":"C# Corner","logo":"https://media.daily.dev/image/upload/t_logo,f_auto/v1/logos/ada2d584df6241748fc4e71878dc70a3","url":"https://daily.dev/sources/csharpcorner"},"commentCount":0,"discussionUrl":"https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx","interactionStatistic":[{"@type":"InteractionCounter","interactionType":{"@type":"LikeAction"},"userInteractionCount":1},{"@type":"InteractionCounter","interactionType":{"@type":"CommentAction"},"userInteractionCount":0}],"keywords":"java,algorithms,binary-search,dynamic-programming","timeRequired":"PT6M"} {"@context":"https://schema.org","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://daily.dev"},{"@type":"ListItem","position":2,"name":"C# Corner","item":"https://daily.dev/sources/csharpcorner"},{"@type":"ListItem","position":3,"name":"Cut Matrix: A Deep Dive into the DP + Binary Search Solution"}]} {"@context":"https://schema.org","@type":"FAQPage","@id":"https://daily.dev/posts/cut-matrix-a-deep-dive-into-the-dp-binary-search-solution-wmrbvf8lx#faq","mainEntity":[{"@type":"Question","name":"How do you reduce the state space for a matrix-cutting DP problem where each cut only leaves the bottom or right piece cuttable?","acceptedAnswer":{"@type":"Answer","text":"The cutting rule restricts the remaining cuttable region to always be a suffix rectangle described by just a starting row r and starting column c, extending to the matrix's bottom-right corner. 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