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Dynamic Programming - Delete Operations For Two Strings #163

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akshitagupta15june merged 1 commit into from
Oct 8, 2022
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Dynamic Programming - Delete Operations For Two Strings #163

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48 changes: 48 additions & 0 deletions DynamicProgramming/DeleteOperationsForTwoStrings.js
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Original file line number Diff line number Diff line change
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/*
Input: word1 = "sea", word2 = "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
*/


function min(i, j){
return i > j ? j : i;
}

function max(i, j){
return i > j ? i : j;
}

function solve(w1, w2, dp, i, j){
// If we have traversed both the strings till the end at the same time,
// that means there are no extra characters, so return 0.
// (Is being handled by the below base case but just added to minimize a max
// operation on some large string sizes)
if(i == w1.length && j == w2.length) return 0;

// If one of the strings is traversed completely before the other,
// that means there are some extra characters in the other string,
// so we add the remaining length to our steps count.
if(i == w1.length || j == w2.length) return max(w1.length-i, w2.length-j);


// return the cached value.
if(dp[i][j] != -1) return dp[i][j];

// if both characters are same that means no need to remove those characters
// from both the strings, so just increment the pointer for both strings.
if(w1[i] == w2[j]) return solve(w1, w2, dp, i + 1, j + 1);

// otherwise we add one to the step count and solve for minimum steps
// from the options of removing from string w1 or w2.
return dp[i][j] = 1 + min(solve(w1, w2, dp, i + 1, j), solve(w1, w2, dp, i, j + 1));
}
function minDistance(word1, word2) {
let dp = Array(word1.length + 1).fill().map(() => Array(word2.length+1).fill(-1))
return solve(word1, word2, dp, 0, 0);
}


let s1 = "sea";
let s2 = "eat";
console.log(minDistance(s1, s2));