Java O(N) time, O(1) space. Best time & space
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(N*10^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(10^2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public boolean isTransformable(String s, String t) {\n        int[][] sr = new int[10][10];\n        int[][] tr = new int[10][10];\n        int[] sc = new int[10];\n        int[] tc = new int[10];\n        boolean equal = true;\n        for (int i = 0; i < s.length(); i++) {\n            int ss = s.charAt(i) - \'0\';\n            int tt = t.charAt(i) - \'0\';\n            if (equal) {\n                switch (Integer.compare(tt,ss)) {\n                    case 1: return false; // cannot produce t > s\n                    case -1: equal = false;\n                }\n            }\n            // running frequencies of digits in s & t\n            sc[ss]++;\n            tc[tt]++;\n            // running inversions in s & t\n            for (int j = 9; j > ss; j--) sr[ss][j] += sc[j];\n            for (int j = 9; j > tt; j--) tr[tt][j] += tc[j];\n        }\n        for (int i = 0; i < 9; i++) {\n            if (sc[i] != tc[i]) return false; // cannot modify digit count\n            for (int j = i+1; j <= 9; j++) {\n                if (tr[i][j] > sr[i][j]) return false;  // cannot add inversions\n            }\n        }\n        return true;\n        \n    }\n}\n```
['Java']